HELP
2. TRIANGLE ABC~TRIANGLE ADE; find x
B
A
20
15
E
2x + 3
C

4√5 is the simplified form of  square root of 10 times square root of 8 .

First, The square root of ten is √10

and, the square root of 8 is √8

Now, simplify square root of 10 times square root of 8.

= √10×√8

By radical property (√a×√b=√a×b

So,√10×√8=√(10×8)=√80

=√16×5

=4√5

Thus, 4√5 is the simplified form of  square root of 10 times square root of 8 .

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(a). 40,320 different arrangements can be made arranging the letters of FABULOUS.

(b). 5,760 different arrangements have the A appearing anywhere before the S (such as in FABULOUS).

(c). 1,440 different arrangements have the first U appearing anywhere before the S (such as in FABU- LOUS).

(d). 120 different arrangements have all four vowels appear consecutively (such as FAUOUBLS).

(a). The word FABULOUS has 8 letters, so there are 8! = 40,320 different arrangements of its letters.

(b). To count the number of arrangements where A appears before S, we can fix A in the first position and S in the last position. Then, we have 6 remaining letters to arrange in the 6 remaining positions. This gives us 6! = 720 possible arrangements where A appears before S.

However, we can also fix A in the second position and S in the last position, and we can fix A in the third position and S in the last position, and so on. Therefore, the total number of arrangements where A appears anywhere before S is 720 * 8 = 5,760.

(c). To count the number of arrangements where the first U appears before S, we can fix U in the first position and S in the last position, and then we have 6 remaining letters to arrange in the 6 remaining positions. This gives us 6! = 720 possible arrangements where the first U appears before S.

However, there are two U's in the word FABULOUS, so we can also fix the second U in the first position and S in the last position, and then we have 6 remaining letters to arrange in the 6 remaining positions. This also gives us 6! = 720 possible arrangements where the first U appears before S. Therefore, the total number of arrangements where the first U appears anywhere before S is 720 * 2 = 1,440.

(d). To count the number of arrangements where all four vowels appear consecutively, we can group the vowels together as one unit, so we have F, B, L, S, and the group AUOU. The group AUOU has 4 letters, so there are 4! = 24 different arrangements of these letters.

However, the group AUOU can appear in any of the 5 positions between F and B, between B and L, between L and S, after S, or before F. Therefore, the total number of arrangements where all four vowels appear consecutively is 24 * 5 = 120.

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To compute the area of the transformed ellipse T(S), A transformed ellipse is an ellipse that has been subjected to a transformation, which can change its size, shape, and orientation.

We'll follow these steps:
1. Identify the original ellipse S and its area.
2. Apply the transformation T(x) = Ax.
3. Compute the determinant of matrix A.
4. Compute the area of the transformed ellipse T(S) using the determinant.

Step 1: The original ellipse S has an area of 11.
Step 2: The transformation T(x) = Ax is given by the matrix A = [1 2; 0 -6].
Step 3: Compute the determinant of matrix A.
The determinant is given by det(A) = (1 * -6) - (2 * 0) = -6.
Step 4: Compute the area of the transformed ellipse T(S) using the determinant.
The area of the transformed ellipse T(S) can be computed using the formula:

Area(T(S)) = |det(A)| * Area(S).
Area(T(S)) = |-6| * 11 = 6 * 11 = 66.

The area of the transformed ellipse T(S) is 66.

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Answer:

Step-by-step explanation:

therefore favourable/total outcomes

hope it helps

There are 24 possible four-letter sequences using the letters b, r, j, and w once each.

To find out how many four-letter sequences are possible using the letters b, r, j, and w once each, we can use the formula for permutations of n objects taken r at a time, which is:

P(n,r) = n! / (n-r)!

In this case, n = 4 (since there are 4 letters to choose from) and r = 4 (since we want to choose all 4 letters). So we can plug in these values and simplify:

P(4,4) = 4! / (4-4)!
P(4,4) = 4! / 0!
P(4,4) = 4 x 3 x 2 x 1 / 1
P(4,4) = 24

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The trigonometric identity [sint + cost]²/sin(t)cos(t) = 2 + csc(t)sec(t).

Trigonometric identities are mathematical equations that contain trigonometric ratios.

Since we have the trigonometric identity

[sint + cost]²/sin(t)cos(t) = 2 + csc(t)sec(t). We want to show that the left-hand-side L.H.S = right-hand-side R.H.S. We proceed as folows

Since we have L.H.S = [sint + cost]²/sin(t)cos(t), so expanding the numerator, we have that

[sint + cost]²/sin(t)cos(t), = [sin²t + 2sintcost + cos²(t)]/sin(t)cos(t)

Using the trigonometric identity sin²t + cos²t = 1, we have that

[sin²t + 2sintcost + cos²(t)]/sin(t)cos(t) =  [sin²t + cos²(t) + 2sintcost]/sin(t)cos(t)

=  [1 + 2sintcost]/sin(t)cos(t)

Dividing through by the denominator sin(t)cos(t) , we have that

[1 + 2sintcost]/sin(t)cos(t) = [1/sin(t)cos(t) + [2sintcost]/sin(t)cos(t)

= 1/sin(t) × `1/cos(t) + 2

= cosec(t)sec(t) + 2  [since cosec(t) = 1/sin(t) and sec(t) = 1/cos(t)]

= 2 +  cosec(t)sec(t)

= R.H.S

Since L.H.S = R.H.S

So, the trigonometric identity [sint + cost]²/sin(t)cos(t) = 2 + csc(t)sec(t).

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The given equation is rearranged to obtain 9x - 1 - (41 - 5x) = 0. By simplifying and solving the equation, we get x = 3.

To solve this equation, we first simplify it by subtracting what is to the right of the equal sign from both sides of the equation. This gives us 9x - 1 - 41 + 5x = 0, which simplifies to 14x - 42 = 0. We then pull out the like factor of 14 to obtain 14(x - 3) = 0.

Next, we use the zero product property to find the values of x that make the equation true. We know that 14 can never equal 0, so we focus on the expression inside the parentheses. Setting x - 3 equal to 0, we get x = 3.

Therefore, the solution to the given equation is x = 3.

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Complete Question:

Simplify 9x-1=41-5x.

c = 1/2, and the pdf is fx(x) = { 1/2 [tex](1-x^{2} )^{-1}[/tex], −1 ≤ x ≤ 1or 0 otherwise and cdf of X is: FX(x) = { 0, x ≤ -1 or -1/2 [tex]tan^{-1}(x/\sqrt{1-x^{2} )}[/tex] + 1/2, -1 < x < 1 or 1 ,x ≥ 1

(a) To find c, we need to integrate the pdf over its support and set the result equal to 1 since the pdf must integrate to 1 over its entire support. Therefore, we have:

1 = ∫c [tex](1-x^{2} )^{-1}[/tex] dx from -1 to 1

Using the substitution u = 1 - [tex]x^{2}[/tex], we have:

1 = c∫ [tex]u^{-1/2}[/tex] dx from 0 to 1

Solving the integral, we get:

1 = 2c

Therefore, c = 1/2, and the pdf is:

fx(x) = {

1/2 [tex](1-x^{2} )^{-1}[/tex] , −1 ≤ x ≤ 1

0 otherwise

To sketch the pdf, we can notice that it is symmetric about x = 0 and that it approaches infinity as x approaches ±1. Therefore, it will have a peak at x = 0 and decrease as we move away from x = 0 in either direction.

(b) To find the cdf of X, we can integrate the pdf from negative infinity to x for each value of x in the support of the pdf. Therefore, we have:

FX(x) = ∫fX(t) dt from -∞ to x

For x ≤ -1, FX(x) = 0, since the pdf is zero for x ≤ -1. For -1 < x < 1, we have:

FX(x) = ∫1/2 [tex](1-t^{2} )^{-1}[/tex] dt from -1 to x

Using the substitution u = [tex]t^{2}[/tex] - 1, we have:

FX(x) = -1/2 ∫[tex](x^{2} -1)^{-1/2}[/tex] du

Solving the integral, we get:

FX(x) =  -1/2 [tex]tan^{-1}(x/\sqrt{1-x^{2} )}[/tex]  + 1/2

For x ≥ 1, we have FX(x) = 1, since the pdf is zero for x ≥ 1. Therefore, the cdf of X is:

FX(x) = {

0 x ≤ -1

-1/2 [tex]tan^{-1}(x/\sqrt{1-x^{2} )}[/tex]  + 1/2 -1 < x < 1

1 x ≥ 1

To sketch the cdf, we can notice that it starts at 0 and increases gradually from x = -1 to x = 1, where it jumps to 1. The cdf is also symmetric about x = 0, similar to the pdf.

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The function f is not a function from r to r since none of the given functions are functions from r to r due to either being undefined for certain inputs or having multiple outputs for a single input.

a) f(x) = 1/x

f is not a function from R to R because it is undefined when x = 0. For f to be a function from R to R, it must have a defined output for every real number input.

b) f(x) = √x

f is not a function from R to R because the square root is not defined for negative numbers. Thus, it doesn't have an output for negative real number inputs.

c) f(x) = ±√(x² + 1)

f is not a function from R to R because it violates the definition of a function, which states that each input must have exactly one output. In this case, each input x has two outputs: the positive and negative square roots of (x² + 1).

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Answer:

Step-by-step explanation:

0.7 cubic kilometers

V= lwh

V=(.5)(2)(.7)

The expression 7 - 3(2x + 5) is equivalent to -6x - 8.

Given the expression in the question:

7 - 3(2x + 5)

We can simplify this expression using the distributive property of multiplication over addition or subtraction.

According to this property, when a number is multiplied by a sum or difference, we can distribute the multiplication over each term within the parentheses.

So, applying the distributive property, we get:

7 - 3(2x + 5)

7 - 3 × 2x -3 × 5

= 7 - 6x - 15

Simplifying further, we can combine like terms:

= -6x - 8

Therefore, the expression to -6x - 8, which is option C.

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Answer: both

Step-by-step explanation:

If the marginal propensity to consume(mpc) = 4/5 then the government purchases multiplier is 5. Therefore, the answer is (d) 5.

The marginal propensity to consume (MPC) is the fraction of each additional unit of income that is spent on consumption. In other words, it is the change in consumption that results from a one-unit change in income.

The government purchases multiplier is a measure of the impact that changes in government purchases of goods and services have on the overall economy. It is the ratio of the change in the equilibrium level of real GDP to the change in government purchases.

The government purchases multiplier (K) is calculated using the formula:

K = 1 / (1 - MPC)

where MPC represents the marginal propensity to consume.

In this case, the MPC is given as 4/5, so we can substitute this value into the formula:

K = 1 / (1 - 4/5)

= 1 / (1/5)

= 5

Therefore, the government puchases multiplier is 5. Therefore, the answer is (d) 5.

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(a) the set of all products in which the exponents sum to 4 is (1 + x + x²)4.

(b) the set of all products in which the exponents sum to 4 is (1 + x + x² + x³ + x⁴)4.

(c) the set of all products in which the exponents sum to 4 is (1 + x³ + x⁴)2(1 + x + x²)2.

(d) the set of all products in which the exponents sum to 4 is (1 + x + x² + x³ ......)3.

An exponent is written as a superscript to the right of the base number and indicates the number of times the base number is multiplied by itself.

(a) The set of all formal products in which the exponents sum to 4 can be determined by the following equation: (1 + x + x²)4.

This is because the exponents of each term must sum to 4, thus the exponents of the terms in the product must also sum to 4. Thus, the set of all products in which the exponents sum to 4 is (1 + x + x²)4.

(b) The set of all formal products in which the exponents sum to 4 can be determined by the following equation: (1 + x + x² + x³ + x⁴)4.

This is because the exponents of each term must sum to 4, thus the exponents of the terms in the product must also sum to 4. Thus, the set of all products in which the exponents sum to 4 is (1 + x + x² + x³ + x⁴)4.

(c) The set of all formal products in which the exponents sum to 4 can be determined by the following equation: (1 + x³ + x⁴)2(1 + x + x²)2.

This is because the exponents of each term must sum to 4, thus the exponents of the terms in the product must also sum to 4. Thus, the set of all products in which the exponents sum to 4 is (1 + x³ + x⁴)2(1 + x + x²)2.

(d) The set of all formal products in which the exponents sum to 4 can be determined by the following equation: (1 + x + x² + x³  ......)3.

This is because the exponents of each term must sum to 4, thus the exponents of the terms in the product must also sum to 4. Thus, the set of all products in which the exponents sum to 4 is (1 + x + x² + x³ ......)3.

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The area under the curve of the function f(x) between the points x = 1 and x = 5 is equal to 6 units.

Given that f is a continuous function and ∫1 5 f(x)dx = 6, we know that the area under the curve of f between 1 and 5 is equal to 6.

We don't have any information about g, so we can't say much about it. However, we can use the fact that f is continuous to make some deductions.

Since f is continuous, we know that it must be integrable on [1, 5]. This means that we can define another function F(x) as the definite integral of f(x) from 1 to x:

F(x) = ∫1 x f(t) dt

We can then apply the fundamental theorem of calculus to F(x) to get:

F'(x) = f(x)

In other words, F(x) is an antiderivative of f(x), so its derivative is f(x).

We can also use the fact that f is continuous to say that it must have an average value on [1, 5]. This average value is given by:

avg(f) = (1/4) ∫1 5 f(x) dx

Using the fact that ∫1 5 f(x)dx = 6, we can simplify this to:

avg(f) = (1/4) * 6 = 1.5

This means that on average, the value of f(x) between 1 and 5 is 1.5.

However, since we don't know anything else about f, we can't say much more than that.


Based on the given information, f and g are continuous functions and the definite integral of f(x) from 1 to 5 is equal to 6. In mathematical notation, it can be expressed as:

∫(from 1 to 5) f(x)dx = 6

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For any m x n matrix A, AAT and ATA are symmetric matrices as  (AAT)^T = AAT and (ATA)^T = ATA.

To prove or disprove that for any m x n matrix A, AAT and ATA are symmetric, we can use the definition of a symmetric matrix and the properties of matrix transposes.
A matrix B is symmetric if B = B^T, where B^T is the transpose of B. So, we need to show that (AAT)^T = AAT and (ATA)^T = ATA.
Step 1: Find the transpose of AAT:
(AAT)^T = (AT)^T * A^T (by the reverse order property of transposes)
(AAT)^T = A * A^T (since (A^T)^T = A)
(AAT)^T = AAT
Step 2: Find the transpose of ATA:
(ATA)^T = A^T * (AT)^T (by the reverse order property of transposes)
(ATA)^T = A^T * A (since (A^T)^T = A)
(ATA)^T = ATA

Since (AAT)^T = AAT and (ATA)^T = ATA, we have proved that for any m x n matrix A, AAT and ATA are symmetric matrices.

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Answer:

Set your calculator to degree mode.

Please draw the figure to confirm my answer.

[tex] \tan( \alpha ) = \frac{15}{24} [/tex]

[tex] \alpha = {tan}^{ - 1} \frac{5}{8} = 32 \: degrees[/tex]

So the angle of elevation is 32°.

D is the correct answer.

Answer:

P=70

Step-by-step explanation:

because when you divide y by x you get 70

140 divided by 2 is 70.

Answer:

**NEED ANSWER ASAP DUE TOMORROW**

What are two observations of quasars that prove they cannot be a part of the milky way galaxy?

Step-by-step explanation:

(a) The ball is 3 feet high when it leaves the child's hand.                      

(b) The maximum height of the ball is 75 feet.

(c) The ball strikes the ground approximately 91.6 feet from the child

(a) To find the height of the ball moving in a projectile motion when it leaves the child's hand, we need to substitute x = 0 into the equation and solve for y:

[tex]y = -1/16(0)^2 + 6(0) + 3[/tex]

y = 3

Therefore, the ball is 3 feet high when it leaves the child's hand.

(b) To find the maximum height of the ball, we need to find the vertex of the parabola defined by the equation. The x-coordinate of the vertex is given by:

x = -b / (2a)

where a = -1/16 and b = 6. Substituting these values, we get:

x = -6 / (2(-1/16)) = 48

The y-coordinate of the vertex is given by:

[tex]y = -1/16(48)^2 + 6(48) + 3 = 75[/tex]

Therefore, the maximum height of the ball is 75 feet.

(c) To find how far from the child the ball strikes the ground, we need to find the value of x when y = 0 (since the ball will be at ground level when its height is 0).

Substituting y = 0 into the equation, we get:

[tex]0 = -1/16x^2 + 6x + 3[/tex]

Multiplying both sides by -16 to eliminate the fraction, we get:

[tex]x^2 - 96x - 48 = 0[/tex]

Using the quadratic formula, we can solve for x:

[tex]x = [96 \ ^+_- \sqrt{(96^2 - 4(-48))}]/2\\x = [96\ ^+_- \sqrt{(9408)}]/2\\x = 48 \ ^+_- \sqrt{(2352)[/tex]

x ≈ 4.4 or x ≈ 91.6

Since the ball is thrown from x = 0, we can discard the negative solution and conclude that the ball strikes the ground approximately 91.6 feet from the child.

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The volume of the solid in the first octant bounded by the parabolic cylinder z=25-x² and the plane y=1 is 32.33 cubic units (approximately).

To find the volume of the solid, we need to integrate the cross-sectional area of the solid with respect to x. Since the plane y=1 cuts the solid into two halves, we can integrate over the range 0 ≤ x ≤ 5.

For a fixed value of x, the cross-sectional area of the solid is given by the area of the ellipse formed by the intersection of the parabolic cylinder and the plane y=1. The equation of this ellipse can be obtained by substituting y=1 in the equation of the parabolic cylinder:

Therefore, the equation of the ellipse is:

We can now integrate the cross-sectional area over the range 0 ≤ x ≤ 5:

where A(x) is the area of the ellipse given by:

Evaluating the integral, we get:

Therefore, the volume of the solid in the first octant bounded by the parabolic cylinder z=25-x² and the plane y=1 is approximately 32.33 cubic units.

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To find the area of the shaded region, we need to subtract the area of the square from the area of the circle:

Area of shaded region = Area of circle - Area of square

Area of shaded region = 81.3 - 16.1

Area of shaded region = 65.2 cm^2

The error interval for the area, a, of the shaded region can be found by considering the errors in the measurements of the areas of the circle and the square. For the area of the circle, the value is rounded to 1 decimal place, so the error is at most 0.05 cm^2 (half of the value of the smallest decimal place). For the area of the square, the value is truncated to 1 decimal place, so the error is at most 0.1 cm^2 (the value of the smallest decimal place).

Thus, the error interval for the area, a, of the shaded region is:

m < a < n

where:

m = 81.3 - 16.1 - 0.1 = 65.1 cm^2

n = 81.3 - 16.1 + 0.05 = 65.25 cm^2

Therefore, the error interval for the area, a, of the shaded region is:

65.1 < a < 65.25

The linear differential operator that annihilates [tex]e^{-x} + 2xe^x - x^2e^x is (D - 2)(D - 1)(D + 1).[/tex]

For [tex]1+6x - 2x^3:[/tex]

The first derivative is [tex]6 - 6x^2[/tex], and the second derivative is -12x. Since the second derivative is a constant multiple of the original function, we can use the differential operator D - 2 to annihilate the function:

[tex](D - 2)(1 + 6x - 2x^3) = D(1 + 6x - 2x^3) - 2(1 + 6x - 2x^3)[/tex]

[tex]= (6 - 6x^2) - 2 - 12x + 4x^3 - 2[/tex]

[tex]= 4x^3 - 6x^2 - 12x + 4[/tex]

[tex]For e^{-x} + 2xe^x - x^2e^x:[/tex]

The first derivative is [tex]2e^x - 2xe^x - x^2e^x,[/tex] and the second derivative is [tex]-2e^x + 2xe^x + 2e^x - 2xe^x - 2xe^x - x^2e^x[/tex]. Simplifying, we get:

[tex](D - 2)(D - 1)(D + 1)(e^{-x} + 2xe^x - x^2e^x) = (D - 1)(D + 1)(2e^x - 2xe^x - x^2e^x)[/tex]

[tex]= (D + 1)(2e^x - 4xe^x - 2xe^x + 2x^2e^x - x^2e^x)[/tex]

[tex]= (D + 1)(2e^x - 6xe^x + 2x^2e^x)[/tex]

Therefore, the linear differential operator that annihilates [tex]e^{-x} + 2xe^x - x^2e^x is (D - 2)(D - 1)(D + 1).[/tex]

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Answer:

the equation is

g=f+10 g=f-10

Answer:

The function represents a growth. The percentage rate is 9%.

Step-by-step explanation:

f(x)= a(1+r)^t

f(x) 7900 (1+.09)^x

Using the above equations, since the rate is greater than 1, that means the function represents growth. And to find the percentage rate, take the .09 and multiply it by 100 to convert it into a percentage.

The thing that I notice about the ratios of the lengths of the intersected segments as you move B and C is that the ratios is altered (changed) as the position of D changes, but the two ratios was one that remain equal to each other.

If triangle and a line goes through two sides of the triangle, the line cuts the sides into littler line sections. When we choose any point on the line, the lengths of the smallest line fragments it makes will change. But, the proportion between the lengths of the littler line portions will continuously remain the same.

For case, in the event that one of the smaller line segments is twice as long as another, at that point this proportion will continuously be 2:1, no matter where we choose the point on the line.

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Answer:

6. From speed vs. time graphs, we know that in order to find the distance, we need to look at the area covered. Given the rectangular shape below, we know that it would be A=lw so A=60mph(2.5h).

7. A=60mph(2.5h)= 150 miles.

(a) The probability that a student does not drink once a month= 0.40

(b) The probability corresponding to each outcome would depend on the           given relative  probability

(c) You can calculate the value of this expression to find the probability    that at least one person drinks once a month.

a. The probability that a student does not drink once a month can be calculated as 1 minus the probability that a student drinks once a month, which is given as P(D) = 0.60.

So, P(not D) = 1 - P(D) = 1 - 0.60 = 0.40.

b. If we randomly select 2 students at a time and record their drinking habits, the sample space corresponding to this experiment would consist of all possible combinations of drinking habits for the two students.

Since each student can either drink once a month (D) or not drink once a month (not D), there are four possible outcomes:

Both students drink once a month (D, D)

Both students do not drink once a month (not D, not D)

The first student drinks once a month and the second student does not (D, not D)

The first student does not drink once a month and the second student drinks once a month (not D, D)

The probability corresponding to each outcome would depend on the given relative frequency or probability of a student drinking once a month (P(D)) and the complement of that probability (1 - P(D)).

c. Assuming that 5 students are selected at random, the probability that at least one person drinks once a month can be calculated using the complement rule.

The complement of the event "at least one person drinks once a month" is "none of the 5 students drink once a month" or "all 5 students do not drink once a month".

The probability that one student does not drink once a month is P(not D) = 0.40 (as calculated in part a).

The probability that all 5 students do not drink once a month is [tex](P(not D))^5[/tex], since the events are assumed to be independent.

So, the probability that at least one person drinks once a month is:

P(at least one person drinks once a month) = 1 - P(none of the 5 students drink once a month)

= 1 - [tex](P(not D))^5[/tex]

= 1 - [tex](0.40)^5[/tex] (after substituting the value of P(not D) from part a)

You can calculate the value of this expression to find the probability that at least one person drinks once a month.

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δy = 0.1192 and dy = -1.08.

To find δy and dy, we can use the formula:
δy = f(x+δx) - f(x)
dy = f'(x) * dx
where f(x) = x^4 and x = -3.

First, let's find δy:
δy = f(-3+0.01) - f(-3)
δy = (-3.01)^4 - (-3)^4
δy = 0.1192

Next, let's find dy:
dy = f'(-3) * 0.01
dy = 4(-3)^3 * 0.01
dy = -1.08

Comparing δy and dy, we can see that they have different signs and magnitudes. δy is positive and has a magnitude of 0.1192, while dy is negative and has a magnitude of 1.08.

This makes sense because δy represents the change in y due to a small change in x, while dy represents the instantaneous rate of change of y with respect to x at a specific point.

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The value of (f^(-1))'(5) is 1/9, where f(x) = x^3 + 9x + 5.

To find the derivative of the inverse function at a particular point, we can use the formula:

(f^(-1))'(y) = 1 / f'(f^(-1)(y))

where y is the value at which we want to find the derivative of the inverse function.

In this case, we want to find (f^(-1))'(5), so we need to first find f'(x) and f^(-1)(5).

Starting with f'(x):

f(x) = x^3 + 9x + 5

Taking the derivative with respect to x, we get:

f'(x) = 3x^2 + 9

Now we need to find f^(-1)(5):

f(x) = 5

x^3 + 9x + 5 = 5

x^3 + 9x = 0

x(x^2 + 9) = 0

x = 0 or x = ±√(-9)

Since the inverse function is a function, it can only take one value for each input value, so we can discard the solution x = -√(-9), which is not a real number. Therefore, f^(-1)(5) = 0.

Now we can use the formula to find (f^(-1))'(5):

(f^(-1))'(5) = 1 / f'(f^(-1)(5)) = 1 / f'(0)

Substituting into the expression for f'(x), we get:

f'(x) = 3x^2 + 9

f'(0) = 3(0)^2 + 9 = 9

(f^(-1))'(5) = 1 / f'(f^(-1)(5)) = 1 / f'(0) = 1 / 9

So, (f^(-1))'(5) = 1/9

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