The engineer wants to modify the roller coaster design by transforming the function. which represents 2 f (0.3 x minus 1) 10, the modified design of the roller coaster?

Considering the definition of perpendicular line, the slope of a line perpendicular to the line whose equation is 3x+y= -9 is  [tex]\frac{1}{3}[/tex].

A linear equation o line can be expressed in the form y = mx + b.

where

Perpendicular lines are lines that intersect at right angles or 90° angles.

Two nonvertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. In other words, if you multiply the slopes of two perpendicular lines, you get –1.

In this case, the line is 3x+y=-9. Expressed in the form y = mx + b, you get: y= -9 - 3x or, what is the same, y= -3x -9

If you multiply the slopes of two perpendicular lines, you get –1. In this case, the line has a slope of  -3. So:

(-3)× slope perpendicular line= -1

slope perpendicular line= (-1)÷ (-3)

slope perpendicular line=  [tex]\frac{1}{3}[/tex]

Finally, the slope of a line perpendicular to the line whose equation is 3x+y= -9 is  [tex]\frac{1}{3}[/tex].

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Answer:

4/10

Step-by-step explanation:

Let the numerator be 2x and the denominator is 5x. The denominator is 6 more than the numerator so if we add 6 to the numerator both numbers should be equal.

Solve:

2x+6 = 5x

6 = 3x

x = 2

2x = 4

5x = 10

Step-by-step explanation:

Let the numerator be 2x and denominator be 5x,

A.T.Q,

2x+6 = 5x

2x-5x = -6

-3x = -6

3x = 6

x = 2

So, the fraction is 4/10 = 2/5

Answer:  2.5 meters

This value is exact without any rounding.

======================================================

Explanation:

The first task is to convert the mixed number 20⅚ into an improper fraction

I'll write 20⅚ as 20 & 5/6

The rule to use is a & b/c = (a*c + b)/c

So,

a & b/c = (a*c + b)/c

20 & 5/6 = (20*6+5)/6

20 & 5/6 = 125/6

This means the volume is exactly (125/6)pi cubic meters.

Plug this into the sphere volume formula and isolate the radius r like so

V = (4/3)*pi*r^3

(125/6)pi = (4/3)*pi*r^3

125/6 = (4/3)*r^3

(4/3)*r^3 = 125/6

4r^3 = 3*(125/6)

4r^3 = 125/2

r^3 = (125/2)*(1/4)

r^3 = 125/8

r = (125/8)^(1/3)

r = 5/2

r = 2.5

Answer:

-456 I think but I could be wrong though

Answer: [tex]\Large\boxed{z=57}[/tex]

Concept:

Here, we need to know how to expand expressions like (a - b)(a + b).

For expressions like (a - b)(a + b), the expanded form must be:

                                                  a² - b²

For checking whether it is true:

 (a - b)(a + b)

= a² - ab + ba - b²

= a² - b²

Solve:

Given equation

[tex](\sqrt{z}+\sqrt{22})(\sqrt{z}-\sqrt22)}=35[/tex]

Expand the parenthesis by the concept

[tex](\sqrt{z})^2-\sqrt{22}^2=35[/tex]

Simplify the exponents

[tex]z-22=35[/tex]

Add 22 on both sides

[tex]z-22+22=35+22[/tex]

[tex]\Large\boxed{z=57}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

The terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

In this question,

The function is f(x) = [tex]\frac{sin(x)}{1-x}[/tex]

The general form of Maclaurin series is

[tex]\sum \limits^\infty_{k:0} \frac{f^{k}(0) }{k!}(x-0)^{k} = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} +\frac{f'''(0)}{3!}x^{3}+......[/tex]

To find the Maclaurin series, let us split the terms as

[tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] ------- (1)

Now, consider f(x) =  sin(x)

Then, the derivatives of f(x) with respect to x, we get

f'(x) = cos(x), f'(0) = 1

f''(x) = -sin(x), f'(0) = 0

f'''(x) = -cos(x), f'(0) = -1

[tex]f^{iv}(x)[/tex] = cos(x), f'(0) = 0

Maclaurin series for sin(x) becomes,

[tex]f(x) = 0 +\frac{1}{1!}x +0+(-\frac{1}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+.....[/tex]

Now, consider [tex]f(x) = (1-x)^{-1}[/tex]

Then, the derivatives of f(x) with respect to x, we get

[tex]f'(x) = (1-x)^{-2}, f'(0) = 1[/tex]

[tex]f''(x) = 2(1-x)^{-3}, f''(0) = 2[/tex]

[tex]f'''(x) = 6(1-x)^{-4}, f'''(0) = 6[/tex]

[tex]f^{iv} (x) = 24(1-x)^{-5}, f^{iv}(0) = 24[/tex]

Maclaurin series for (1-x)^-1 becomes,

[tex]f(x) = 1 +\frac{1}{1!}x +\frac{2}{2!}x^{2} +(\frac{6}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=1+x+x^{2} +x^{3} +......[/tex]

Thus the Maclaurin series for [tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] is

⇒ [tex]f(x)=(x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+..... )(1+x+x^{2} +x^{3} +......)[/tex]

⇒ [tex]f(x)=x+x^{2} +x^{3} - \frac{x^{3} }{6} +x^{4}-\frac{x^{4} }{6} +.....[/tex]

⇒ [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex]

Hence we can conclude that the terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

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Rejected masses x are those that weigh less than 9.02 oz or more than 9.48 oz.

To find the rejected masses:

Therefore, rejected masses x are those that weigh less than 9.02 oz or more than 9.48 oz.

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The correct question is given below:

A company did a quality check on all the packs of trail mix it manufactured. each pack of trail mix is targeted to weigh 9.25 oz. a pack must weigh within 0.23 oz of the target weight to be accepted. what is the range of rejected masses, ×, for the manufactured trail mixes?

Answer:

The length of the diagonal of a rectangle is 10m

Step-by-step explanation:

There is a visual below,

The diagonal of a rectangle has formed 2 right triangles.

This diagonal is called hypotenuse side length of a triangle

To find the hypotenuse, we can use the Pythagorean Theorem

a² + b² = c²

we are given the length of a and b, where a = 6 and b = 8

(6)² + (8)² = c²

36 + 64 = c²

100 = c²

√100 = c

10 = c

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Answer:

6x³ + 14x² + 6x - 12

Step-by-step explanation:

(3x - 2)(2x² + 6x + 6)

6x³ - 4x² + 18x² - 12x + 18x - 12

6x³ + 14x² + 6x - 12

Answer: 6x³+14x²+6x-12

Step-by-step explanation:

Here, we can use the distributive property to multiply one of the factors by each term in the other. Below, I multiplied [tex]2x^2 + 6x + 6[/tex] by both 3x and -2.

[tex](2x^2+ 6x + 6) (3x - 2)\\3x(2x^2+ 6x + 6)-2(2x^2+ 6x + 6)\\6x^3+18x^2+18x-4x^2-12x-12[/tex]

Now, we can rearrange so that like terms are next to each other. Then, we can combine them to simplify.

[tex]6x^3+18x^2+18x-4x^2-12x-12\\6x^3+18x^2-4x^2+18x-12x-12\\6x^3+14x^2+6x-12[/tex]

Answer:

109°

Step-by-step explanation:

The sum of angles in all triangles is equal to 180. For this given triangle, we can solve for x.

48° + 23° + x° = 180°
180° - 71° = x°
x = 109°

The product of [tex](x - 8)^2[/tex] is not a polynomial equation because a polynomial equation means a coefficient of x that has a power of two or more powers.

The equations developed with variables, exponents, and coefficients exists named polynomial equations. It can include various exponents, where the higher one stands named the degree of the equation.

Closure property under multiplication notes that any two rational numbers' outcome will be a rational number, i.e. if a and b exist in any two rational numbers, ab will also be a rational number.

Example: (3/2) × (2/9) = 1/3.

The product of [tex]$(x - 8)^2[/tex]

Simplifying the equation as (x − 8)(x − 8)

Apply Perfect Square Formula, and we get

[tex]$(a-b)^{2}=a^{2}-2 a b+b^{2}$[/tex]

[tex]$&(x-8)^{2}=x^{2}-2 x \cdot 8+8^{2} \\[/tex]

[tex]$&=x^{2}-2 x \cdot 8+8^{2}[/tex]

Simplifying the above equation, we get

[tex]$x^{2}-2 x \cdot 8+8^{2}=x^{2}-16 x+64 \\[/tex]

[tex]$&=x^{2}-16 x+64[/tex]

This is not a polynomial equation because a polynomial equation means a coefficient of x that has a power of two or more powers.

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The standard error of the sample mean to the nearest hundredth  is  mathematically given as

S.E=0.784

The amount by which the population means deviates from a sample mean is represented by the standard error of the mean, which is more often referred to as simply the standard error.

It informs you how much the sample means would change if you were to perform an experiment using fresh samples from the same population but this time uses different samples.

The standard error is obtained by calculating the standard deviation and then dividing that number by the square root of the sample size. It determines the accuracy of a sample mean by factoring in the variation in sample means that exists from one set of data to the next.

Generally, The population's mean height is 62.95 inches, so let's use that.

Taking into account that the population has a standard deviation of 5.65 inches

The sample mean is used to calculate the standard error of the mean.

[tex]SE=\frac{SD}{\sqrt{n}}[/tex]

Therefore

[tex]SE=\frac{5.65}{ \sqrt{52}}[/tex]

S.E=0.7835

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Answer:

9/4

Step-by-step explanation:

We follow bodmas

4/3 +( -1/6 + 13/12)

( lcm = 12)

( -2+ 13/12)

( 11/ 12)

4/3 + ( 11/12)

4/3 + 11/12

lcm = 12 also

and that will equal to

=16 + 11/ 12

= 27/ 12

divide by 3 to simplest form

= 9/4

Answer:

First and second angles going upward

Answer: 2, 4, 6

Step-by-step explanation:

Answer:

Step-by-step explanation:

1. yes

2. 0.048955

3. equal, not equal, less than and equal to, less than, greater than and equal to, greater than

4. 9 51/250

5. 387.5%

Answer:

b

Step-by-step explanation:

Answer:

B. Monday

Step-by-step explanation:

Monday, Tuesday, Wednesday, Monday, Tuesday, Wednesday, Monday

hope it helps

Answer:

11.25

77.85

Step-by-step explanation:

1.6x + 3 = 21

use inverse operation (+ = -, x = ÷) by taking 3 to the opposite side and changing it's integer to negative

1.6x = 21 - 3

divide 2 sides by 1.6

1.6x = 18

1.6x/1.6 = 18/1.6

x = 11.25//

2.15(x-5)=75

inverse operation

x= 75 - 2.15 + 5

x= 77.85//

Answer:

[tex]\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x[/tex]

Rewrite 25 as 5²:

[tex]\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x[/tex]

Integration by substitution

[tex]\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}[/tex]

[tex]\textsf{Let }x=5 \sin \theta[/tex]

[tex]\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}[/tex]

Find the derivative of x and rewrite it so that dx is on its own:

[tex]\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta[/tex]

[tex]\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta[/tex]

Substitute everything into the original integral:

[tex]\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}[/tex]

Take out the constant:

[tex]\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta[/tex]

[tex]\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}[/tex]

[tex]\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

[tex]\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :[/tex]

[tex]\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

Take out the common factor 1/2:

[tex]\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

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Answer:

30°

Step-by-step explanation:

To make sure it's complete

The area of the region cleaned by the 7-in wiper blade with the angle subtended being 92⁰ as described in the task content is; 39.32in².

According to the task content, it follows that the wiper blade is 7-in long and subtends an angle of 92°.

Consequently, the area swept by the blade in discuss can be determined by the Area of sector formula and hence, is;

Area = (92/360) × (22/7) × 7².

Area = 39.32 in².

Ultimately, it follows from the solving steps above that the area of the region cleaned by the 7-in wiper blade with the angle subtended being, 92⁰ is; Area = 39.32 in².

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The mean, median, and mode all have the same values in a symmetrical distribution. The mean is frequently chosen as the primary indicator of central tendency in these situations.

The mean, median, and mode frequently occur at the same location in a symmetrical distribution, where the values of variables exist at regular frequencies. The graph's middle can be divided into two sides that mirror one another by drawing a line through it.

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Using a quadratic regression equation, it is found that the prediction of the number of songs that Kimberly will download on Month 9 is of:

c. 16.

To find the equation, we need to insert the points (x,y) in the calculator.

In this problem, we have that the function initially decreases, then it increases , which means that it is a quadratic function. The points (x,y) to  be inserted into the calculator are given as follows:

(1,8), (2,6), (4,5), (6,7), (7,9)

Inserting these points into the calculator, the prediction of y for a value of x is given as follows:

y = 0.391x² - 2.937x + 10.455

Hence, when x = 9, the prediction is:

y = 0.391 x 9² - 2.937 x 9 + 10.455

y = 15.69

Rounded to 16, hence option c is correct.

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Answer:

25.12 rounded.

Step-by-step explanation:

C = 2[tex]\pi[/tex]r

C=[tex]\pi[/tex]d

C = [tex]\pi[/tex](8)

C = 25.12 rounded.

Answer:

25 and 1/7 mm

Step-by-step explanation:

Diameter = circunference / π

π = 3.1416    aprox.

Then:

8 = citcunference / 3.1416

8 * 3.1416 = circunference

25.1428 = circunference

1/7 = 0.1428

Then:

25.1428 = 25 + 1/7

= 25 1/7

Answer:

Let , the constant factor be x so 3 X + 5 x + 7 x = 180

Step-by-step explanation:

Find out the value of x

7x+3x+5x=180

15x=180

x=180/15

x=12

now,

     x=3 multiplied by 12=36

     x=7 multiplied by 12=84

      x=5 multiplied by 12=60

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Type I error says that we suppose that the null hypothesis exists rejected when in reality the null hypothesis was actually true.

Type II error says that we suppose that the null hypothesis exists taken when in fact the null hypothesis stood actually false.

In statistics, a Type I error exists as a false positive conclusion, while a Type II error exists as a false negative conclusion.

Making a statistical conclusion still applies uncertainties, so the risks of creating these errors exist unavoidable in hypothesis testing.

The probability of creating a Type I error exists at the significance level, or alpha (α), while the probability of making a Type II error exists at beta (β). These risks can be minimized through careful planning in your analysis design.

Examples of Type I and Type II error

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To fill in the values we have the following

a. log6⁰, log 2 . 1, log 8/3

b. log 1 .4 , 1 , log 4. 6

c. log 9/2, log 3. 5 , log 5⁷

To do this, you have to decide on that particular number that you want to find the logarithm on. Next you have to find the base of that number.

The logarithm of the number is the power that it would have to be raised for us to obtain a different number. You have to note that the logarithm of the number is the exponent that a base would have to be raised up to in order to get a particular number.

Log6⁰ for instance would give us the solution of 1 as the answer. While telling us that we have that the exponent is 0 while the base is 6.

One good property of logarithm is that log m/n = log m - log n also when we have log mn, it is the same as log m * log n

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Answer:

K(6, 9)

Step-by-step explanation:

Let the K coordinates be (x, y)

Mid-point formula:

[tex]\sf (x_m, y_m) = (\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2})[/tex]

Applying formula:

[tex]\sf (-3,\:-1) = (\:\dfrac{-12+x}{2} ,\: \dfrac{-11+y}{2} )[/tex]

Comparing expression:

[tex]\sf \dfrac{-12+x}{2} = -3 \quad and \quad \dfrac{-11+y}{2} = -1[/tex]

[tex]\sf -12+x = 2(-3) \quad and \quad -11+y = 2(-1)[/tex]

[tex]\sf x = -6 + 12 \quad and \quad y = -2 + 11[/tex]

[tex]\sf x = 6 \quad and \quad y = 9[/tex]

So, coordinates of K is (6, 9)