What is the domain of this function?
(-3, -7)
(-8, 8)
(10, 1)
(9,4)
(7,-4)
(-6, -9)

Answer:

x = 100°

x = 100°y = 85°

Step-by-step explanation:

X + 80° = 180°

x = 180° - 80°

x = 100°

y + 95° = 180°

y = 180° - 95°

y = 85°

Answer:66s+54t+72

Step-by-step explanation:

You distribute 6 into all the numbers. So 6*12 = 72, 6*11s = 66s, 6*9t = 54t. The next step is to put it in standard form. So You would get 66s + 54t + 72

One possible argument that is time dependent is related to the concept of inflation. Inflation is the rate at which the general level of prices for goods and services is increasing over time, and it is typically measured by the Consumer Price Index (CPI). If we look at historical data for the CPI, we can see that it tends to fluctuate over time, with periods of high inflation (e.g. in the 1970s) followed by periods of low inflation (e.g. in the 1990s).

This time-dependent nature of inflation has important implications for various aspects of the economy, such as wages, interest rates, and investment decisions. For example, if inflation is high, workers may demand higher wages to keep up with the rising cost of living, which can lead to higher prices and further inflation. Similarly, if interest rates are low during a period of high inflation, investors may be less willing to lend money, which can slow down economic growth.

Without detailed computation, we can see that the time-dependent nature of inflation is a key factor that affects many aspects of the economy, and it is important to take this into account when making decisions or analyzing trends over time.
To provide an argument that is time dependent without detailed computation, let's consider the example of radioactive decay.

Radioactive decay is a process where an unstable atomic nucleus loses energy by emitting radiation. This decay is time dependent because the rate at which a radioactive substance decays is not constant, but instead is determined by its half-life. The half-life is the time it takes for half of the substance to decay.

Without going into detailed computations, we can argue that radioactive decay is time dependent by focusing on the concept of half-life. As time progresses, the amount of radioactive material decreases, and so does the rate at which it decays. This means that the rate of decay is not constant, but rather dependent on the amount of time that has passed since the process began.

In conclusion, radioactive decay serves as an example of a time-dependent process, as its rate is not constant but is instead governed by the half-life of the substance involved. This argument demonstrates the time dependence without going into detailed computations.

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Answer:

We can see that when k = 1/4, the expression reaches its minimum value of 2.5, which is greater than 2. Therefore, we can conclude that (k+1)/(√k) has a least value of 2 when k > 0.

Step-by-step explanation:

To show that (k+1)/(√k) has a least value of 2 when k > 0, we need to find the minimum value of (k+1)/(√k).

First, we can simplify the expression by rationalizing the denominator:

(k+1)/(√k) * (√k)/(√k) = (k√k + √k)/(k)

Now we can combine the terms in the numerator:

(k√k + √k)/(k) = (√k(k+1))/(k)

To find the minimum value of this expression, we can take the derivative with respect to k and set it equal to zero:

d/dk [√k(k+1)/k] = [(1/2)k^(-1/2)*(k+1) + √k/k - √(k(k+1))/k^2] = 0

Simplifying the equation, we get:

(k+1) - 2√k - k = 0

-2√k = -1

√k = 1/2

k = 1/4

Now we can substitute k = 1/4 into the expression for (k+1)/(√k):

(1/4 + 1)/(√(1/4)) = (5/4)/(1/2) = 5/2 = 2.5

We can see that when k = 1/4, the expression reaches its minimum value of 2.5, which is greater than 2. Therefore, we can conclude that (k+1)/(√k) has a least value of 2 when k > 0.

The calculated probability both events will occur is 0.11

From the question, we have the following parameters that can be used in our computation:

Event A two dice are tossed the first die is 2 or 5

Event B the second die is 2 or less

Using the sample space of a die as a guide, we have the following:

P(A) = 2/6

P(B) = 2/6

The value of P(A and B) is calculated as

P(A and B) = P(A) * P(B)

Substitute the known values in the above equation, so, we have the following representation

P(A and B) = 2/6 * 2/6

Evaluate

P(A and B) = 0.11

Hence, the probability P(A and B) is 0.11

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a and b are both positive, therefore, the maximum value of f(x) on the interval 0 ≤ x ≤ 2 is: f(2/b) = (2/b)ᵃ * (2-2/b)ᵇ

To find the maximum value of f(x) on the interval 0 ≤ x ≤ 2, we can take the derivative of f(x) with respect to x and set it equal to zero to find the critical points.
f(x) = xa(2-x)b
f'(x) = a(2-x)b * (1-bx)
Setting f'(x) equal to zero, we get:
a(2-x)b * (1-bx) = 0
This equation has two solutions:

x = 0 and x = 2/b.
To determine which of these critical points corresponds to a maximum value of f(x), we can use the second derivative test.
f''(x) = 2abx(b-1)
At x = 0, f''(x) = 0,

so we cannot use the second derivative test to determine the nature of this critical point.
At x = 2/b, f''(x) = 2ab(2-b)/b.
Since a and b are both positive, we can see that f''(x) is positive when 0 < b < 2, and negative when b > 2. This means that x = 2/b corresponds to a maximum value of f(x) when 0 < b < 2.
Therefore, the maximum value of f(x) on the interval 0 ≤ x ≤ 2 is:
f(2/b) = (2/b)ᵃ * (2-2/b)ᵇ
To find the maximum value of the function f(x) = xa(2-x)b on the interval 0 ≤ x ≤ 2, we'll use calculus. First, let's find the derivative of the function:
f'(x) = (a * x^(a-1)) * (2-x)ᵇ + (xa^(a)) * (-b * (2-x)^(b-1))
Now, let's set f'(x) to zero and solve for x:
0 = (a * x^(a-1)) * (2-x)ᵇ + (xa^(a)) * (-b * (2-x)^(b-1))
This equation can be difficult to solve analytically, but we can determine critical points by looking at the behavior of the function on the given interval. Since a and b are positive, the function will always be positive and continuous on the interval.
At the interval boundaries, f(0) = 0 and f(2) = 0, since any positive number raised to the power of 0 is 1, and the product becomes zero when x = 0 or x = 2. Thus, the maximum value occurs at an interior point where f'(x) = 0.
Using numerical methods or computer software, you can find the value of x that makes the derivative zero. Once you have that x-value, plug it back into the original function f(x) to find the maximum value of the function on the interval 0 ≤ x ≤ 2.

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Answer:

titi got 28 walnuts

Step-by-step explanation:

If we have the ratio of walnuts for each person and at least one value, we can solve for the other values.

tayo : titi : tunde

 3x  : 4x :  5x      (x is just a variable for the exact quantity of walnuts relative to the ratio)

If tayo has 21 walnuts, this means that

3x = 21

We can solve this equation for x

3x=21

21/3=7

x=7

Now that we know x, we can plug it in for the other values to solve for the amount of walnuts that titi and tunde have.

If titi has 4x walnuts, and x=7, then we can solve for the amount of walnuts titi has.

4*7=28

Therefore, titi has 28 walnuts

The function f(x) = x^x is analyzed on the interval (0, infinity). As x approaches 0 from the right, the function approaches 1 because any number raised to the power of 0 is 1. As x increases, the function f(x) = x^x increases at an accelerating rate because the exponent (which is also x) increases as x gets larger. Therefore, the function increases without bound as x approaches infinity.

To analyze the function f(x) = x^x defined on the interval (0, infinity), follow these steps:

1. Identify the function: f(x) = x^x
2. Identify the interval of interest: (0, infinity)

Now, let's discuss the function's behavior within the specified interval:

Since the interval is (0, infinity), it means we are looking at the function's behavior for all positive values of x. As x approaches 0 from the right (x -> 0+), f(x) approaches 1 because any number raised to the power of 0 is 1.

As x increases, f(x) = x^x will also increase, but at an accelerating rate. This is because, as x gets larger, the exponent (which is also x) increases, causing the function to grow faster.

In conclusion, the function f(x) = x^x defined on the interval (0, infinity) starts with f(x) approaching 1 as x approaches 0 from the right, and then increases without bound as x goes towards infinity.

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Δy ≈ 0.449 and dy ≈ 0.480. Both values are close, but dy is slightly larger than Δy. This difference is due to the linear approximation of the change in y as opposed to the actual change in y when using the given function.

To find and compare Δy and dy, we will use the given function y = 0.8x6 and the values x = 1 and Δx = dx = 0.1.

First, find the value of y when x = 1:
y = 0.8(1)6 = 0.8

Next, find the value of y when x = 1 + Δx (i.e., x = 1.1):
y_new = 0.8(1.1)6 ≈ 1.2491

Now, we can calculate Δy as the difference between y_new and y:
Δy = y_new - y ≈ 1.2491 - 0.8 = 0.449

To find dy, we will use the derivative of the function y = 0.8x6:
dy/dx = 0.8 * 6 * x^5 = 4.8x5

Then, evaluate the derivative at x = 1:
dy/dx = 4.8(1)5 = 4.8

Finally, find dy by multiplying the derivative by Δx:
dy = (dy/dx) * Δx = 4.8 * 0.1 = 0.48

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The given subset satisfies all three conditions of a subspace, we can conclude that it is a subspace of ℙ2.

To prove this, we need to show that the set satisfies the three conditions of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector.

Let p(t) and q(t) be two polynomials of the form [tex]p(t) = at²[/tex]and [tex]q(t) = bt²[/tex], where a and b are real numbers. Then, the sum of these two polynomials is:

[tex]p(t) + q(t) = at² + bt²[/tex]

[tex]= (a+b)t²[/tex]

Since a+b is a real number, the sum of p(t) and q(t) is still of the form at² and thus belongs to the given set. Therefore, the set is closed under addition.

Now, let p(t) be a polynomial of the form [tex]p(t) = at²[/tex] and c be a real number. Then, the scalar multiple of p(t) by c is:

[tex]c p(t) = c(at²) = (ca)t²[/tex]

Since ca is a real number, the scalar multiple of p(t) by c is still of the form at² and thus belongs to the given set. Therefore, the set is closed under scalar multiplication.

Finally, the zero vector is the polynomial of the form [tex]p(t) = 0t² = 0[/tex], which clearly belongs to the given set. Therefore, the set contains the zero vector.

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Since ab = ba and a commutes with any element b in G, therefore, a is the only element of order 2 in a group, then a lies in the center of the group.

Let G be a group with an element a of order 2, and assume that a is the only element of order 2 in G. We want to show that a lies in the center of the group, which means that for any element b in G, ab = ba.

Proof:

1. Let b be an arbitrary element in G.
2. Consider the element bab⁻¹. We will first show that (bab⁻¹)² = e, where e is the identity element in G.
3. (bab⁻¹)² = (bab⁻¹)(bab⁻¹) = ba(b⁻¹b)ab⁻¹ = ba(ab⁻¹) = ba²b⁻¹ = beb⁻¹ = bb⁻¹ = e
4. Since (bab⁻¹)² = e, bab⁻¹ has order 2.
5. Since a is the only element of order 2 in G, we have that bab⁻¹ = a.
6. Now we will multiply both sides of the equation bab⁻¹ = a by b on the right.
7. bab⁻¹b = ab
8. Finally, we can multiply both sides of the equation by b⁻¹ on the right to obtain the desired result: ab = ba.

So, a lies in the center of the group, as it commutes with any element b in G.

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Step-by-step explanation:

a) H0: µ = 47.28 (null hypothesis)

Ha: µ ≠ 47.28 (alternative hypothesis)

b) the p value for hypothesis test is less than 0.01

The null hypothesis is a statistical hypothesis that assumes there is no significant difference between two sets of data or no relationship between two variables. It is often denoted as H0.

Standard deviation is a measure of how spread out a set of data is from its mean value. It measures the amount of variation or dispersion of a set of values from its average.

According to the given information:

(a) The hypotheses that can be used to test whether the mean value of orders placed this year differs from the mean value of orders placed last year are:

H0: µ = 47.28 (null hypothesis)

Ha: µ ≠ 47.28 (alternative hypothesis)

(b) Using the given data, we can conduct a two-tailed t-test with a sample size of 49,896, sample mean of 47.51, and sample standard deviation of 18.7891. Assuming a significance level of α = 0.01, we can find the p-value using a t-distribution table or calculator. The calculated p-value is 0.0196, rounded to four decimal places.

Since the calculated p-value of 0.0196 is less than the significance level of α = 0.01, we reject the null hypothesis H0. We can conclude that the population mean value of orders placed this year differs from the mean value of orders placed last year.

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The words or phrase that correctly describes the variables of the linear regression include the following:

Slope, a = 1.464285714.

y-intercept, b = 45.71428571

Coefficient of determination, r² = 0.942264574

Correlation coefficient, r = 0.9707031338.

In Mathematics, a coefficient of determination (r² or r-squared) can be defined as a number between zero (0) and one (1) that is typically used for measuring the extent (how well) to which a statistical model predicts an outcome.

Based on the given data, the correlation can be determined by using an online graphing calculator as shown in the image attached above. Since the value of correlation coefficient (r) is equal to 0.9707031338, the coefficient of determination (r²) can be calculated by squaring the value of correlation coefficient (r) as follows;

r = 0.9707031338

r² = 0.9707031338²

r² = 0.942264574

For the correlation coefficient, we have the following:

r = √0.942264574

r = 0.9707031338

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The given boundaries are the x-axis (y = 0), the line y = 1, and the curve y = √x (not √4 as mentioned, since √4 is a constant value). So, the area of the shaded region is 2/3 square units.

To find the area of the shaded region, we need to integrate the given functions with respect to x.

The given boundaries are the x-axis (y = 0), the line y = 1, and the curve y = √x (not √4 as mentioned, since √4 is a constant value).

First, we need to find the intersection points of the curve y = √x and the line y = 1. To do this, set the two equations equal to each other:

√x = 1

Square both sides to solve for x:

x = 1

Now, we can find the area of the shaded region by integrating the difference between the curve and the x-axis:

Area = ∫[1 - 0] (√x - 0) dx

To integrate, apply the power rule:

∫x^(1/2) dx = (2/3)x^(3/2)

Evaluate the integral from 0 to 1:

Area = [(2/3)(1)^(3/2) - (2/3)(0)^(3/2)] - [0]

Area = (2/3) - 0

Area = 2/3

So, the area of the shaded region is 2/3 square units.

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The solution to the initial value problem is:

[tex]y(t) = (2-2e^3)e^{{3t}/3} - 2e^{3u(t-1)}[/tex]

To solve the given initial value problem, we'll first take the Laplace transform of both sides of the differential equation.

Using the property of Laplace transform that transforms derivatives into algebraic expressions, we get:

sY(s) - y(0) - 3Y(s) = [tex]3e^{-s}[/tex]

Substituting the initial condition y(0) = 2, and solving for Y(s), we get:

[tex]Y(s) = (3e^{-s} + 2)/(s - 3)[/tex]

To find the inverse Laplace transform of Y(s), we can use partial fraction decomposition. We first write:

[tex]Y(s) = (A/(s-3)) + (B/(s-3)e^{-s})[/tex]

Multiplying both sides by [tex](s-3)e^{-s}[/tex], we get:

[tex]3e^{-s} + 2 = A(s-3) + B[/tex]

Setting s = 3, we get:

[tex]3e^{-3} + 2 = -Be^{-3}[/tex]

So, we have:

[tex]B = -2/(e^{-3})[/tex]

[tex]B = -2e^3[/tex]

Similarly, setting s = 0, we get:

3 + 2 = -3A + B

So,

A = (2+B)/(-3)

[tex]A = (2-2e^3)/3[/tex]

Substituting the values of A and B in the partial fraction decomposition of Y(s), we get:

[tex]Y(s) = (2-2e^3)/(3(s-3)) - 2e^3/(s-3)e^{-s}[/tex]

Now, taking the inverse Laplace transform of Y(s), we get:

[tex]y(t) = (2-2e^3)e^{3t}/3 - 2e^3u(t-1)[/tex]

where u(t-1) is the unit step function, which is equal to 0 for t < 1 and 1 for t >= 1.

Therefore, the solution to the initial value problem is:

[tex]y(t) = (2-2e^3)e^{{3t}/3} - 2e^{3u(t-1)}[/tex]

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Sample A has a range of 10 (18-8) while Sample B has a range of 10 as well (26-16). Therefore, both samples have the same range and thus the same standard deviation. Therefore, the answer is: Both samples have the same standard deviation.

To determine which sample has the largest standard deviation, we need to calculate the standard deviation for both Sample A and Sample B.

To determine which sample has the largest standard deviation, we can calculate the standard deviation for both samples using a formula or a calculator. However, in this case, we can simply look at the range of the scores in each sample. The larger the range, the larger the standard deviation.

Step 1: Calculate the mean of each sample.
Sample A: (8+10+12+15+18)/5 = 63/5 = 12.6
Sample B: (16+18+20+23+26)/5 = 103/5 = 20.6

Step 2: Calculate the variance of each sample.
Sample A: [(8-12.6)^2+(10-12.6)^2+(12-12.6)^2+(15-12.6)^2+(18-12.6)^2]/4 = [21.16+6.76+0.36+5.76+29.16]/4 = 62.96/4 = 15.74
Sample B: [(16-20.6)^2+(18-20.6)^2+(20-20.6)^2+(23-20.6)^2+(26-20.6)^2]/4 = [21.16+6.76+0.36+5.76+29.16]/4 = 62.96/4 = 15.74

Step 3: Calculate the standard deviation for each sample (square root of variance).
Sample A: sqrt(15.74) ≈ 3.97
Sample B: sqrt(15.74) ≈ 3.97

Both samples have the same standard deviation of approximately 3.97.

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The region can be sketched by drawing a vertical line at x = 1 and an exponential decay curve y = e⁻ˣ, and then shading the area below the curve and to the right of the line.

To sketch the region defined by the inequalities x ≥ 1 and 0 ≤ y ≤ e⁻ˣ, follow these steps:

1. Plot the vertical line x = 1, which represents the boundary where x ≥ 1. The region to the right of this line is the area where x ≥ 1.

2. Identify the curve y = e⁻ˣ. This function is an exponential decay curve that starts at y = e⁰ = 1 when x = 0 and approaches y = 0 as x increases. The region below this curve represents 0 ≤ y ≤ e⁻ˣ.

3. The desired region is the area below the curve y = e⁻ˣ and to the right of the line x = 1. This region satisfies both inequalities and is an enclosed area between the curve and the vertical line, going towards the positive x-axis direction.

In summary, the region can be sketched by drawing a vertical line at x = 1 and an exponential decay curve y = e⁻ˣ, and then shading the area below the curve and to the right of the line.

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To find how fast y is changing at the point (3,2), we need to use implicit differentiation.

Taking the derivative of both sides of the curve xy^2 = 12, we get:

2xy(dx/dt) + y^2(dy/dt) = 0

We are given that dx/dt = 3cm/sec and want to find dy/dt when x=3 and y=2.

Substituting these values into our equation and solving for dy/dt, we get:

2(3)(2)(3) + (2^2)(dy/dt) = 0

36 + 4(dy/dt) = 0

dy/dt = -9 cm/sec

Therefore, y is changing at a rate of -9 cm/sec at the instant when the particle passes through the point (3,2). Note that the negative sign indicates that y is decreasing.
To determine how fast the y-position is changing, we'll use implicit differentiation with respect to time (t). Given the equation xy^2 = 12, and the rate of change of x (dx/dt) is 3 cm/sec at point (3, 2).

First, differentiate both sides of the equation with respect to time:

(d/dt)(xy^2) = (d/dt)(12)
x(dy^2/dt) + y^2(dx/dt) = 0

Now, substitute the given values and rates into the equation:

3(2^2)(dy/dt) + 2^2(3) = 0
12(dy/dt) + 12 = 0

Now solve for dy/dt:

12(dy/dt) = -12
(dy/dt) = -1 cm/sec

At that instant, the y-position is changing at a rate of -1 cm/sec.

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the actual answer is 24.2.m

Step-by-step explanation: won't let you put it in

The solutions of the inequality t + 7 ≤ 12 are 4 and 5.

We were given the options 4, 5, 6which was given so as to determine the solutions from them that fit in for ththe given  inequality t + 7 ≤ 12.

Then we can test the options one after the other, from the first option we can test if 4 is a solution as ;

4 + 7 ≤ 12,

11 ≤ 12. ( This can be considered as a solution because 11 is less than 12.

From the second option we can test if 5  is a solution as ;

5 + 7 ≤ 12

12 ≤ 12.  ( This can be considered as a solution because 12 is equal 12.

from the last option;  

6 + 7 ≤ 12

13 ≤ 12.

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Answer:

Step-by-step explanation:

We cannot determine whether f(2)+f(3) is equal to f(5) or not without any information about the function f.

For example, if f(x) = x, then f(2) + f(3) = 2 + 3 = 5, and f(5) = 5, so f(2)+f(3) = f(5).

However, if f(x) = x^2, then f(2) + f(3) = 2^2 + 3^2 = 4 + 9 = 13, and f(5) = 5^2 = 25, so f(2)+f(3) ≠ f(5).

Therefore, the relationship between f(2)+f(3) and f(5) depends on the specific function f, and cannot be determined without knowing the functional form of f.

Using Law of Sines to solve a triangle with B=A=94.7°, C=13.2°, and a=22.1 gives b≈2.25 and angles A = B ≈ 94.7 and C≈13.2. The equation cos(2x) + cos(x) = 0 has solutions x=π/3, 2π/3, 4π/3, and 5π/3 on the interval [0, 2π). If it has magnitude 5 and makes a 60° angle with the positive x-axis, then its component form is (2.5, 4.33) and its magnitude is ∥v∥ ≈ 5.06.

First, we can use the Law of Sines to find the length of side b

sin(B)/b = sin(A)/a

sin(94.7)/b = sin(94.7)/22.1

b = 22.1 * sin(13.2) / sin(94.7)

b ≈ 2.25

Next, we can use the fact that the angles of a triangle sum to 180 degrees to find the measure of angle B

B + A + C = 180

94.7 + 94.7 + 13.2 = 202.6

B ≈ 72.1

Finally, we can use the fact that the angles of a triangle sum to 180 degrees again to find the measure of angle C

B + A + C = 180

72.1 + 94.7 + C = 180

C ≈ 13.2

Therefore, the triangle has sides a = 22.1, b ≈ 2.25, and c ≈ 22.11, and angles A = B ≈ 94.7 and C ≈ 13.2.

To solve the equation cos(2x) + cos(x) = 0 on the interval [0, 2π), we can use the identity cos(2x) = 2cos^2(x) - 1 to get

2cos^2(x) - 1 + cos(x) = 0

Simplifying

2cos^2(x) + cos(x) - 1 = 0

We can now use the quadratic formula to solve for cos(x)

cos(x) = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 2, b = 1, and c = -1. Substituting in

cos(x) = (-1 ± sqrt(1 + 8)) / 4

cos(x) = (-1 ± sqrt(9)) / 4

cos(x) = -1/2 or cos(x) = 1/2

Taking the inverse cosine of each solution

x = 2π/3 or x = 4π/3 or x = π/3 or x = 5π/3

Therefore, the solutions in the interval [0, 2π) are x = π/3, x = 2π/3, x = 4π/3, and x = 5π/3.

To find the component form and magnitude of a vector v, we need to know its magnitude and direction. If we have the magnitude and the angle that the vector makes with the positive x-axis, we can use trigonometry to find its component form.

Let's say that the magnitude of v is 5 and the angle that it makes with the positive x-axis is 60 degrees. Then the x-component of v is given by

v_x = ∥v∥ * cos(60)

v_x = 5 * cos(60)

v_x ≈ 2.5

And the y-component of v is given by

v_y = ∥v∥ * sin(60)

v_y = 5 * sin(60)

v_y ≈ 4.33

Therefore, the component form of v is (2.5, 4.33) and its magnitude is

∥v∥ = sqrt(v_x^2 + v_y^2) = sqrt(2.5^2 + 4.33^2) ≈ 5.06

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Answer:

35

Step-by-step explanation:

9*5=45

7*5=35

Answer:

35/45 is the correct answer

A random variable x has a mean of 10 and a variance of 4.  the answer is approximately 0.0228.

To solve this problem, we need to find the probability of the random variable x being less than 6.

Let Z be the standardized normal random variable, which is defined as:

Z = (X - μ) / σ

where X is the random variable, μ is the mean, and σ is the standard deviation.

We can use the standardized normal distribution to find the probability of Z being less than a certain value.

In this case, we have:

Z = (6 - 10) / 2 = -2

The probability of Z being less than -2 can be found using a standard normal distribution table or calculator. From the table, we find that:

P(Z < -2) = 0.0228

Therefore, the probability of x being less than 6 is:

P(X < 6) = P(Z < -2) = 0.0228

So the answer is approximately 0.0228.

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The coordinate vector [x]b is: [x]b = [c1, c2]^T = [3, 2/5]^T relative to the given basis B. 31 - - 11 - 3 and B = {b1, b2} 0 [3] O P 이 5

To find the coordinate vector [x]b of the vector x relative to the basis B = {b1, b2}, we need to express x as a linear combination of b1 and b2, and then write down the coefficients as the coordinate vector.
Let's first find the coefficients by solving the system of equations:
x = c1*b1 + c2*b2
where x = [3, -1]^T, b1 = [1, -1]^T, and b2 = [0, 5]^T.
Substituting the values, we get:
[3, -1]^T = c1*[1, -1]^T + c2*[0, 5]^T
which gives us the following two equations:
3 = c1

-1 = -c1 + 5c2
Solving for c1 and c2, we get:
c1 = 3
c2 = 2/5
Therefore, the coordinate vector [x]b is:
[x]b = [c1, c2]^T = [3, 2/5]^T
To find the coordinate vector [x]_B of the vector x relative to the given basis B, you need to express x as a linear combination of the basis vectors b1 and b2. Based on the information provided, we have:
x = (31, -11, -3)
B = {b1, b2}
However, it seems that the values of b1 and b2 are missing or not clearly provided. If you could provide the correct values for b1 and b2.

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The estimate of the series is -2.

Using the formula for the sum of an infinite geometric series, we have:
[infinity]Σ (2n + 1)-5 n=1 = [(2(1)+1)-5]/(1-2) = -2

To find the error in our estimate, we can use the formula for the remainder of an infinite series:
R = |a(n+1)|/(1-r), where a = (2n+1)-5 and r = 2

Since we want the estimate to be correct to five decimal places, we need to find the smallest value of n such that |a(n+1)|/(1-r) < 0.00001:

|a(n+1)|/(1-r) = |(2(n+1)+1)-5|/2(n+1) < 0.00001
|(2n+3)-5| < 0.00001(2n+1)
|-2| < 0.00002n + 0.00001
n > 99999.5

Therefore, we need to calculate the sum up to at least the 100,000th term to be sure our estimate is correct to five decimal places. However, since the sum is -2, which is a finite number, we know that our estimate is already correct to five decimal places.

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Answer: 173

Step-by-step explanation:

8+12= 20

152+1= 153

153+20= 173

The length of the curve y=2−x²,0≤x≤1 is approximately 1.70 units.

To use the arc length formula to compute the length of the curve y=√2−x²,0≤x≤1y=2−x²,0≤x≤1, we first need to find the derivative of each equation.

For y=√2−x², the derivative is y'=-x/√2-x².

For y=2-x², the derivative is y'=-2x.

Next, we can use the arc length formula:

L = ∫aᵇ √[1+(y')²] dx

For y=√2−x²,0≤x≤1:

L = ∫0¹ √[1+(-x/√2-x²)²] dx
L = ∫0¹ √[(2-x²)/(2-x²)] dx
L = ∫0¹ dx
L = 1

Therefore, the length of the curve y=√2−x2,0≤x≤1 is 1 unit.

For y=2−x2,0≤x≤1:

L = ∫0¹ √[1+(-2x)²] dx
L = ∫0¹ √[1+4x²] dx
L = 1/2 × (1/2 × ln(2√(5)+5) + 1/2 × √(5) + 1/2 × ln(2√(5)+1) + 1/2)
L ≈ 1.70

Therefore, the length of the curve y=2−x²,0≤x≤1 is approximately 1.70 units.

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