(a) find the differential d y . y = tan x d y = incorrect

The smallest possible value of nullity(a) is 0, and the largest possible value of nullity(a) is 4. To determine the minimum and maximum possible values of nullity(a) for a 7 × 4 matrix, we need to consider the properties of a matrix and nullity.

Nullity(a) is defined as the dimension of the null space of the matrix 'a'. It is also equal to the number of linearly independent columns in the matrix that are not part of its column space.
Since the matrix is a 7 × 4 matrix, it has 4 columns. The rank-nullity theorem states that:
rank(a) + nullity(a) = number of columns in matrix 'a'
The minimum possible value of nullity(a) occurs when all columns are linearly independent, which would mean the rank of the matrix is at its maximum value. In this case, the maximum rank of a 7 × 4 matrix is 4. So, the smallest possible value of nullity(a) is: nullity(a) = number of columns - rank(a)
nullity(a) = 4 - 4 = 0
The maximum possible value of nullity(a) occurs when the rank of the matrix is at its minimum value. In this case, the minimum rank of a 7 × 4 matrix is 0. So, the largest possible value of nullity(a) is: nullity(a) = number of columns - rank(a)
nullity(a) = 4 - 0 = 4
To summarize, the smallest possible value of nullity(a) is 0, and the largest possible value of nullity(a) is 4.

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Answer:

# of quarters: 12 # of dimes: 7

# of large boxes:  14 # of small boxes: 12

Step-by-step explanation:

Problem 1: Dimes and Quarters

Let q represent the number of quarters
Let d represent the number of dimes

We have as a first equation
q + d = 19  (1)

Each quarter is worth $0.25 so q quarters worth = 0.25q

Each dime is worth $0.10 so d dimes worth = 0.10d = 0.1d

Total value of q quarters and d dimes
0.25q + 0.1d = 3.70 (2)

We have two equations in 2 variables which can be solved as follows

Verify:
12 x 0.25 + 7 x 0.10 = 3.70

Problem 2: Apricots

Solution strategy is the same as Problem 1 so I am skipping lengthy explanations

Let L be the number of large boxes, S be the number of small boxes

We have
L + S = 26   (1)

Total cost of boxes

7L + 4S = 146  (2)

Multiply equation (1) by 4 :
4L + 4S = 4 x 26 = 104  (3)

Subtract (2) from (3):
7L + 4S - (4L + 4S) = 146 - 104
3L = 42
L = 42/3 = 14

Substitute in (1)
14 + S = 26
S = 26-14
S = 12

So there are 14 large and 12 small boxes

Check:
14 x 7 + 12 x 4 = 146

Fido's share of the dog food as a percentage, is 36.4%.

Percentage is a way of expressing a number as a fraction of 100. It is represented by the symbol "%". It is a useful way of expressing proportions, ratios, and rates.

First, we need to find the fraction of dog food that Engle gets.

Let x be the fraction of dog food that Engle gets. Then, the fraction of dog food that Fido gets is 1 - x (since Engle and Fido share the rest of the food).

We know that the ratio of Engle's food to Fido's food is 7:4, which means that:

x / (1 - x) = 7/4

Solving for x:

4x = 7(1 - x)

11x = 7

x = 7/11

So Engle gets 7/11 of the dog food, and Fido gets 1 - 7/11 = 4/11 of the dog food.

Now we need to find Fido's share as a percentage:

Fido's share = (4/11) * 100% = 36.4%

Therefore, Fido's share of the dog food is 36.4%.

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The desired formula for the area of the region bounded by the curve, the line, and the perpendiculars to the line from P and Q is

Area of region = ∫p^q 2π[f(x) - mx - b]√[1 + m²] dx

In this project, we are asked to discover formulas for finding the volume of a solid of revolution and the area of a surface of revolution when the axis of rotation is a slanted line. We need to show that the area of the region bounded by the curve y = f(x), the line y = mx + b, and the perpendiculars to the line from P and Q is given by a particular formula.

To find the area of the region bounded by the curve y = f(x), the line y = mx + b, and the perpendiculars to the line from P and Q, we can approximate the area using rectangles perpendicular to the line. The figure given in the problem can be used to express Δu in terms of Δx.

Let us consider a small segment of the curve y = f(x) between two points x and x + Δx. The length of this segment is approximately given by √(Δx² + Δy²), where Δy is the change in y between the two points.

The perpendicular distance between the line y = mx + b and this segment is given by |mx + b - f(x)|. Therefore, the area of the rectangle formed by this segment and the line y = mx + b is approximately given by [√(Δx² + Δy²)][|mx + b - f(x)|][Δx].

Summing up the areas of all such rectangles over the length of the curve gives us an approximation for the area of the region bounded by the curve, the line, and the perpendiculars to the line from P and Q. As we take finer and finer rectangles, this approximation gets closer and closer to the true area.

Taking the limit as the width of the rectangles approaches zero gives us the exact area of the region. Simplifying the expression and substituting the values of P, Q, and the line y = mx + b, we get the formula:

Area of region = ∫p^q 2π[f(x) - mx - b]√[1 + m²] dx

This is the desired formula for the area of the region bounded by the curve, the line, and the perpendiculars to the line from P and Q.

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Answer:

If there is not a lot of rain and each bell pepper grows to 8 ounces, then the total weight of 270 bell peppers would be:

270 x 8 = 2160 ounces or 135 pounds (since 16 ounces = 1 pound).

If there is a lot of rain and each bell pepper grows to 14 ounces, then the total weight of 270 bell peppers would be:

270 x 14 = 3780 ounces or 236.25 pounds (since 16 ounces = 1 pound)

However, based on the two situations above, it is likely that if there is a lot of rain, the total weight of the bell peppers will be greater than 270 pounds, and lower than 270 pounds, respectively.

No matter how much experience you have, good traders never stop learning.

Quite often, we have to learn the same lessons over and over, yet we will still make the same mistakes. It is important to constantly remind ourselves of key principles.

Last week I covered five of the 10 most important lessons I've learned in my 25-year career as a trader.

These lessons are:

In fact, Losing trades are just part of the process of finding winning trades. There is no shame in picking a stock that turns out to be a dud. If you don't have losing trades, then you probably are not taking sufficient risk. The only way you can produce great returns is to accept the fact that there is risk and that a trade may not work.

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The two vector functions that parametrize the intersection of the surfaces are:

[tex]r1(t) = (t, t, (\sqrt{(1+4t^2)+t^2-7)/2} )\\r2(t) = (t, t, -\sqrt{(t^2+64)} )[/tex]

To parametrize the intersection of the surfaces y^2−z^2=z−7 and y^2−z^2=64 using t=y as the parameter, we can first solve for y^2−z^2 in terms of y.

From the equation y^2−z^2=z−7, we get y^2−z^2−z+7=0.

Using the quadratic formula, we can solve for z:

[tex]z = (±\sqrt{(1+4y^2)+y^2-7)/2} )[/tex]

Similarly, from the equation y^2−z^2=64, we get z^2−y^2=-64. Solving for z in terms of y, we get:

[tex]z = ±\sqrt{(y^2+64)}[/tex]

Now we can use t=y as the parameter, and write:

x = t
y = t
[tex]z =±\sqrt{(1+4t^2)+t^2-7)/2}[/tex],

if y^2−z^2=z−7
[tex]z = ±\sqrt{t^2+64)}[/tex]

if y^2−z^2=64

So the two vector functions that parametrize the intersection of the surfaces are:
[tex]r1(t) = (t, t, (\sqrt{(1+4t^2)+t^2-7)/2} )\\\\r2(t) = (t, t, -\sqrt{(t^2+64)} )[/tex]

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Rob received $3.75

First, find out how much the 2 sandwiches cost:

2 sandwiches at $2.50 each

$2.50 × 2 = $5.00

1 bottle of water is $1.25

You add the items bought, the sandwiches and the bottle of water, to find out how much was spent

$5.00 + $1.25 = $6.25

Now, to find out how much change Rob received, you subtract the amount purchased to the amount of the bill he paid with.

$10.00 - $6.25 = $3.75

Answer:

B- nearly symmetrical

Step-by-step explanation:

it’s not perfect, and it’s not skewed in either direction so it leaves b, nearly symmetry

So the probability of drawing a card that is a queen or a club is 4/13.

There are 4 queens and 13 clubs in a standard deck of 52 playing cards. However, the queen of clubs is counted in both groups, so we need to subtract it once. Therefore, the total number of cards that are either a queen or a club (excluding the queen of clubs) is 4 + 13 - 1 = 16.

The probability of drawing a card that is either a queen or a club is the number of desired outcomes (16) divided by the total number of possible outcomes (52):

P(queen or club) = 16/52 = 4/13

So the probability of drawing a card that is a queen or a club is 4/13.

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a. This statement is true. We can write b = ak and d = cl for some integers k and l. Then, b + d = ak + cl = a(k + l). Since a|b, we know a must divide ak and thus a also divides ak + cl. Therefore, a + d = a(k + l) is divisible by a.

b. This statement is also true. Similar to part a, we can write b = ak and d = cl for some integers k and l. Then, bd = akcl = (ac)(kl). Since a|b and c|d, we know that a and c must divide ak and cl respectively. Therefore, ac must divide akcl = bd.

c. This statement is false. Consider a = 2, b = 4, c = 2, and d = 6. We have a|b and b|c, but a does not divide c. Therefore, the statement does not hold for all integers a, b, c, and d.

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a. This statement is true. We can write b = ak and d = cl for some integers k and l. Then, b + d = ak + cl = a(k + l). Since a|b, we know a must divide ak and thus a also divides ak + cl. Therefore, a + d = a(k + l) is divisible by a.

b. This statement is also true. Similar to part a, we can write b = ak and d = cl for some integers k and l. Then, bd = akcl = (ac)(kl). Since a|b and c|d, we know that a and c must divide ak and cl respectively. Therefore, ac must divide akcl = bd.

c. This statement is false. Consider a = 2, b = 4, c = 2, and d = 6. We have a|b and b|c, but a does not divide c. Therefore, the statement does not hold for all integers a, b, c, and d.

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The Laplace transform of a bt + ct^2 for some constants a, b, and c can be found using the linearity property of the Laplace transform. The Laplace transform of a linear combination of functions is equal to the linear combination of their Laplace transforms. Therefore, the Laplace transform of a bt + ct^2 is equal to the Laplace transform of at + the Laplace transform of bt^2.

The Laplace transform of at is a/s, and the Laplace transform of bt^2 is 2b/s^3. Therefore, the Laplace transform of a bt + ct^2 is:

a/s + 2b/s^3

This is the direct answer to the problem.

In more detail, the Laplace transform is a mathematical tool that allows us to convert a function of time into a function of complex frequency. It is defined as the integral of the function multiplied by the exponential function e^(-st), where s is the complex frequency parameter. The Laplace transform has many applications in engineering, physics, and mathematics, particularly in the analysis of linear time-invariant systems.

In this problem, we used the linearity property of the Laplace transform to find the Laplace transform of a bt + ct^2. This property states that the Laplace transform of a linear combination of functions is equal to the linear combination of their Laplace transforms. We first found the Laplace transform of at and bt^2 separately using the Laplace transform formulas. Then, we added them together to obtain the Laplace transform of a bt + ct^2.

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The Laplace transform of a bt + ct^2 for some constants a, b, and c can be found using the linearity property of the Laplace transform. The Laplace transform of a linear combination of functions is equal to the linear combination of their Laplace transforms. Therefore, the Laplace transform of a bt + ct^2 is equal to the Laplace transform of at + the Laplace transform of bt^2.

The Laplace transform of at is a/s, and the Laplace transform of bt^2 is 2b/s^3. Therefore, the Laplace transform of a bt + ct^2 is:

a/s + 2b/s^3

This is the direct answer to the problem.

In more detail, the Laplace transform is a mathematical tool that allows us to convert a function of time into a function of complex frequency. It is defined as the integral of the function multiplied by the exponential function e^(-st), where s is the complex frequency parameter. The Laplace transform has many applications in engineering, physics, and mathematics, particularly in the analysis of linear time-invariant systems.

In this problem, we used the linearity property of the Laplace transform to find the Laplace transform of a bt + ct^2. This property states that the Laplace transform of a linear combination of functions is equal to the linear combination of their Laplace transforms. We first found the Laplace transform of at and bt^2 separately using the Laplace transform formulas. Then, we added them together to obtain the Laplace transform of a bt + ct^2.

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It means that if the calculated z-score is less than -2.17, we can reject the null hypothesis and conclude that the population mean is less than 16 oz with 0.015 level of significance.

To find the associated critical value for a one-tailed z-test with the hypothesis Hₐ: μ < 16 oz and a significance level of 0.015, follow these steps:

1. Identify the type of test: Since the hypothesis states that the mean is less than 16 oz, this is a one-tailed test (left-tailed).

2. Determine the significance level: The significance level is given as 0.015.

3. Find the critical value: Using ahttps://brainly.com/question/13776238 (z) distribution table or a calculator, look up the value that corresponds to the given significance level. Since this is a left-tailed test, we will look for the z-value that has 0.015 of the area to the left.

Upon checking a z-table or calculator, the critical value is approximately -2.17.

So, the associated critical value for this test is -2.17.

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Answer:

Step-by-step explanation:

The determinant is the price of related goods, specifically the price of fuel. The increase in fuel prices would lead to a decrease in the demand for Ford F150 trucks, shifting the demand curve to the left.

The price of related goods is an important determinant of demand.

The drastic increase in fuel prices is likely to shift the demand curve for Ford F150 trucks to the left.

This is because consumers tend to reduce their demand for vehicles with lower fuel efficiency when fuel prices increase.

The determinant of this shift in demand is the price of related goods, specifically, the price of gasoline.

When the price of gasoline increases, it affects the demand for vehicles that are less fuel-efficient, such as trucks.

Therefore, the price of related goods is an important determinant of demand.

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To make brass with 25 ounces of copper, one requires 20.4 ounces of zinc to maintain the required ratio in order to make brass, an alloy that is 55% copper and 45% zinc by weight.

According to the question,

The weight percentage of copper in brass = 55%

The weight percentage of zinc in brass = 45%

Given, the weight of copper = 25 ounces

Let the mass of brass be x

We can say, 55% of x is 25 ounces

[tex]\frac{55}{100}* x= 25\\ \\x=\frac{25*100}{55} \\\\x=\frac{500}{11}[/tex]

Weight of zinc in this alloy = 45% of x

[tex]= \frac{45}{100}*\frac{500}{11}\\\\ =\frac{45*5}{11}=20.45[/tex]

Therefore, the weight of zinc is 20.4 ounces.

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The limit of lim x→/4 5 − 5 tan(x) sin(x) − cos(x) is -5.

To find the limit of lim x→/4 5 − 5 tan(x) sin(x) − cos(x), we can use algebraic manipulation and trigonometric identities.

First, we notice that as x approaches 4, tan(x) approaches infinity and sin(x) and cos(x) approach 0. So, we have an indeterminate form of infinity times 0.

To simplify this expression, we can use the trigonometric identity sin(x)cos(x) = 1/2 sin(2x).

So, we have: lim x→/4 5 − 5 tan(x) (sin(x) − cos(x)sin(x)cos(x)) = lim x→/4 5 − 5 tan(x) (sin(x) − 1/2 sin(2x)) = lim x→/4 5 − 5 tan(x) sin(x) + 5/2 tan(x) sin(2x) = 5 − 5 (1/0) + 5/2 (1/0)

Since we have an indeterminate form of infinity minus infinity, we can't directly evaluate the limit. However, we can use L'Hopital's rule to take the derivative of the numerator and denominator separately and evaluate the limit again.

Taking the derivative of the numerator, we get: -5 sec^2(x) sin(x) + 5 cos(x) Taking the derivative of the denominator, we get: 1

So, applying L'Hopital's rule, we have:

lim x→/4 (-5 sec^2(x) sin(x) + 5 cos(x)) / (cos(x))

= lim x→/4 (-5 sin(x)/cos^2(x) + 5 cos(x)/cos(x))

= lim x→/4 (-5 tan(x)/cos(x) + 5) = -5

Therefore, the limit of lim x→/4 5 − 5 tan(x) sin(x) − cos(x) is -5.

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The expected payoff from this game is $1.25 when you roll a standard 6-sided die two times and get paid the LOWER of the two rolls.

There are 6 × 6 = 36 equally likely outcomes when rolling two dice, where each outcome is determined by the values shown on each die. Since the lower of the two rolls is paid only if the rolls are different, there are 6 outcomes where the two rolls are the same (i.e., a pair of 1s, a pair of 2s, and so on), and 30 outcomes where the two rolls are different.

To find the expected payoff from this game, we need to consider the possible payouts and their corresponding probabilities. Since the lower of the two rolls is paid, the possible payouts range from $1 to $6.

For example, if the two rolls are a 1 and a 4, then the lower roll is 1 and the payout is $1. If the two rolls are a 5 and a 6, then the lower roll is 5 and the payout is $5.

For each possible payout, we need to calculate the probability of it occurring. For a payout of $1, there are 5 outcomes where the two rolls are different and the lower roll is 1, so the probability is 5/36.

For a payout of $2, there are 4 outcomes where the two rolls are different and the lower roll is 2, so the probability is 4/36. Continuing in this way, we can calculate the probabilities for each possible payout as shown in the table below:

Payout Probability

$1 5/36

$2 4/36

$3 3/36

$4 2/36

$5 1/36

$6 0

To find the expected payoff, we need to multiply each payout by its corresponding probability and then add up the results:

E(X) = $1 × 5/36 + $2 × 4/36 + $3 × 3/36 + $4 × 2/36 + $5 × 1/36 + $0 × 6/36

= $5/36 + $8/36 + $9/36 + $8/36 + $5/36 + $0

= $1.25

Therefore, the expected payoff from this game is $1.25.

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Answer:

27.5%

Step-by-step explanation:

Let's denote the total marks of the exam as 'T'.

We know that Tarang got 25% marks and failed by 64 marks, so we can write an equation:

0.25T - 64 = 0 (since he failed)

Solving for T, we get:

T = 256

We also know that if Tarang had got 40% marks, he would have secured 32 marks more than the pass marks. So we can write another equation:

0.4T - Pass marks = 32

Substituting T = 256, we get:

0.4(256) - Pass marks = 32

102.4 - Pass marks = 32

Pass marks = 70.4

Therefore, to pass the exam, Tarang needs to get at least 70.4/T * 100% = 27.5% marks (rounded to one decimal place).

Answer: -7/10

Step-by-step explanation:

if the number is x then x-10=8x

move terms to one side so 7x = -10

divide by 7 to get x by itself so x= -7/10

The p-value for testing the association between PSA and CAPSULE variation by ethnicity is 0.0231. For Black patients, the odds ratio for tumor penetration with a 10-unit PSA increase is 1.6. For White patients, the odds ratio for tumor penetration with a 10-unit PSA increase is 1.3.

The p-value for the LRT can be obtained from the difference in deviance between the two models and comparing it to a chi-squared distribution with 1 degree of freedom. From the given output, the full model has a deviance of 328.12 and the reduced model has a deviance of 329.42. The difference in deviance is 1.3, and the corresponding p-value is 0.253. Therefore, the p-value the researchers should use is 0.253.

To find the estimated odds ratio of tumor penetration associated with a ten unit increase in PSA for Black patients, we look at the coefficient for the interaction term between PSA and Ethnicity (Black) in the output, which is 0.0396. The odds ratio is given by exp(β), where β is the coefficient. So, exp(0.0396*10) = 1.43, which means that for Black patients, a ten unit increase in PSA is associated with a 43% increase in the odds of tumor penetration, controlling for digital rectal results and Gleason scores.

Similarly, to find the estimated odds ratio for White patients, we can look at the coefficient for the main effect of PSA (since White is the reference category for Ethnicity) in the output, which is 0.0593. So, exp(0.0593*10) = 1.78, which means that for White patients, a ten unit increase in PSA is associated with a 78% increase in the odds of tumor penetration, controlling for digital rectal results and Gleason scores.

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Answer:

Publix gives out approximately 12,890.5 cookies, with a margin of error of 1745.1.

Step-by-step explanation:

The survey of 100 store managers found that the average number of cookies given out on a weekday was 15.5, with a margin of error of 2.1.

This means that the true average number of cookies given out on a weekday at all Publix locations in Florida is estimated to be between 13.4 and 17.6, with 95% confidence.

Since Publix has 831 locations in Florida, we can estimate the total number of cookies given out on a weekday by multiplying the average number of cookies by the number of locations:

Total number of cookies = 15.5 * 831 = 12,890.5

Since this is just an estimate based on a sample of 100 stores, there is some uncertainty in this number. The margin of error for the total number of cookies can be calculated using the margin of error for the sample mean and the formula:

Margin of error for total number of cookies = Margin of error for sample mean * Number of locations

Using the margin of error of 2.1, we get:

Margin of error for total number of cookies = 2.1 * 831 = 1745.1

Therefore, we can estimate that on a weekday, Publix gives out approximately 12,890.5 cookies, with a margin of error of 1745.1.

Hope this helps!

Answer:

Publix gives out approximately 12,890.5 cookies, with a margin of error of 1745.1.

Step-by-step explanation:

The survey of 100 store managers found that the average number of cookies given out on a weekday was 15.5, with a margin of error of 2.1.

This means that the true average number of cookies given out on a weekday at all Publix locations in Florida is estimated to be between 13.4 and 17.6, with 95% confidence.

Since Publix has 831 locations in Florida, we can estimate the total number of cookies given out on a weekday by multiplying the average number of cookies by the number of locations:

Total number of cookies = 15.5 * 831 = 12,890.5

Since this is just an estimate based on a sample of 100 stores, there is some uncertainty in this number. The margin of error for the total number of cookies can be calculated using the margin of error for the sample mean and the formula:

Margin of error for total number of cookies = Margin of error for sample mean * Number of locations

Using the margin of error of 2.1, we get:

Margin of error for total number of cookies = 2.1 * 831 = 1745.1

Therefore, we can estimate that on a weekday, Publix gives out approximately 12,890.5 cookies, with a margin of error of 1745.1.

Hope this helps!

Answer:

The answer for x is 37° to the nearest whole number

Answer:

the answer of X is 37° to the nearest whole number

The set can be written as: {6a + 2b | a, b ∈ Z} which means that the elements of the set are all possible values of 6a + 2b where a and b are integers. Written with braces, the set would look like this: { ..., -14, -8, -2, 4, 10, 16, ... }.

This set is defined as the set of all possible values obtained by multiplying any integer 'a' by 6 and any integer 'b' by 2 and then adding the results. Therefore, the set contains all the integers that can be expressed as 6a + 2b, where 'a' and 'b' are integers.

To list the elements of this set between braces, we need to find all possible combinations of values of 'a' and 'b' that belong to the set of integers 'Z' and substitute them into the expression 6a + 2b. This gives us the set of all possible values that can be obtained by multiplying an integer by 6 and another integer by 2 and then adding the results.

The resulting set includes all even integers since 6a is even for all integer values of a, and the sum of two even integers is always even. Therefore, we can write the set as follows:

{..., -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, ...}

Understanding how to define and list the elements of sets is essential in various fields, including mathematics, statistics, and computer science. It allows us to organize data, study relationships between variables, and make predictions or decisions based on the properties of the set.

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A parameterization of curves in the xy-plane requires us to provide a function in terms of a parameter θ, which is usually the angle associated with a point on the curve. The parameters need to be adjusted according to the size and center of the curve, as well as the direction of tracing.

A circle is a two-dimensional shape representing a round, enclosed area with a curved line, or perimeter, usually drawn with a pencil or pen. A circle has no beginning or end and is composed of a single line that forms a continuous loop. A circle is often referred to as a round shape, though its shape is technically an ellipse. A circle's area is determined by its radius, which is the distance from the center of the circle to its perimeter. The circumference of a circle is the total length of its perimeter. Circles are found everywhere in nature and have been used in various forms of art and design throughout history.

1. A circle of radius 3 centered at the origin and traced out clockwise:
x = 3cosθ, y = 3sinθ, where 0 ≤ θ ≤ 2π.

2. A circle of radius 5 centered at the point (2, 1) and traced out counterclockwise:
x = 2 + 5cosθ, y = 1 + 5sinθ, where 2π ≤ θ ≤ 0.

In conclusion, a parameterization of curves in the xy-plane requires us to provide a function in terms of a parameter θ, which is usually the angle associated with a point on the curve. The parameters need to be adjusted according to the size and center of the curve, as well as the direction of tracing.

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[tex]\begin{cases} (x-1)^2-(x+2)^2=9y\\\\ (y-3)^2-(y+2)^2=5x \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-1)^2-(x+2)^2=9y\implies (x^2-2x+1)-(x^2+4x+4)=9y \\\\\\ (x^2-2x+1)-x^2-4x-4=9y\implies -6x-3=9y\implies -3(2x+1)=9y \\\\\\ 2x+1=\cfrac{9y}{-3}\implies 2x+1=-3y\implies 2x=-3y-1\implies x=\cfrac{-3y-1}{2} \\\\[-0.35em] ~\dotfill\\\\ (y-3)^2-(y+2)^2=5x\implies (y^2-6y+9)-(y^2+4y+4)=5x \\\\\\ (y^2-6y+9)-y^2-4y-4=5x\implies -10y+5=5x[/tex]

[tex]\stackrel{\textit{substituting from above}}{-10y+5=5\left( \cfrac{-3y-1}{2} \right)}\implies -10y+5=\cfrac{-15y-5}{2} \\\\\\ -20y+10=-15y-5\implies 10=5y-5\implies 15=5y \\\\\\ \cfrac{15}{5}=y\implies \boxed{3=y} \\\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{-3y-1}{2}}\implies x=\cfrac{-3(3)-1}{2}\implies \boxed{x=-5}[/tex]