The force F (in pounds) needed on a wrench handle to loosen a certain bolt varies inversely with the length L (in inches) of the handle. A force of 40 pounds is needed when the handle is 7 inches long. If a person needs 20 pounds of force to loosen the bolt, estimate the length of the wrench handle. Round answer to two decimal places if necessary.
in inches

Answer:

I think the choose (B)

5x/x + 3/x

Answer:

I thinkchoose no.3

5x+3

5x+3x

Answer:

in a square all sides are equal so x has to equal

128

Hope This Helps!!!

Answer:

Rs. 100 and Rs. 150

Step-by-step explanation:

the ratio = 2:3 => the sum = 2+3=5

Sita gets = 2/5 × 250 = 100

Ram gets = 3/5 × 250 = 150

Answer:

1. Find the GCF of all the terms in the polynomial.

2. Express each term as a product of the GCF and another factor.

3. Use the distributive property to factor out the GCF.

Answer: (5) -5 > -18​

Step-by-step explanation:

-5 is 5 away from zero, making it larger than -18 since it's closer to 0, while -18 is 18 away from zero.

Answer:

b) - 3

Step-by-step explanation:

If x = -3 , then

x + 3 = -3 + 3 = 0

So, denominator would become 0. So , anything divided by 0 is undefined

Answer:

87

Step-by-step explanation:

Given is a figure of cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary.

Therefore,

z° + 93° = 180°

z° = 180° - 93°

z° = 87°

z = 87

Answer:

Addition equation = -4-0) + [(-13)-(-4)]

Answer =  -13

Step-by-step explanation:

For the small arrow in the diagram, the expression is (-4 - 0)

For the bog arrow, the expression will be -13 - (-4)

Adding both expressions

Addition = (-4-0) + [(-13)-(-4)]

Addition = (-4) + (-13+4)

Addition = -4 + (-9)

Addition = -4-9

Addition = -13

9514 1404 393

Answer:

  x = -1, 1, 9

Step-by-step explanation:

You want x such that f(x) = g(x). Subtracting g(x) from both sides of this equation lets us rewrite it as ...

  f(x) -g(x) = 0

  x³ -9x² -(x -9) = 0

  x²(x -9) -1(x -9) = 0 . . . factor the first pair of terms

  (x² -1)(x -9) = 0 . . . . . . use the distributive property

  (x -1)(x +1)(x -9) = 0 . . . . factor the difference of squares

Values of x that make these factors zero will make f(x) = g(x):

  x = -1, 1, 9 for f(x) = g(x)

Answer:

g(x) = x^3 - 2

Step-by-step explanation:

As you can see on the graph, the line has been translated down 2 units.

If the graph of f has the same shape as the graph of g, then the slope remains the same. The y intercept (k) changes by -2 units, so the k value is -2

g(x) = x^3 - 2

Hope this helps!!

Answer:

The answer is:

[tex]H_0: p=0.83\\\\H_a: p \neq 0.83[/tex]

Step-by-step explanation:

Now, we're going to test if sociologists claim to be have visited a region of 0.83 by a person picked randomly on Time In New York City.

Therefore, null or other hypotheses are:

[tex]H_0: p=0.83\\\\H_a: p \neq 0.83[/tex]

Answer:

21

Step-by-step explanation:

7+7+7=21

Answer:

100 psia

Step-by-step explanation:

Applying,

Pressure law,

P/T = P'/T'................. Equation 1

Where P = initial pressure, P' = Final pressure, T = initial temperature, P' = Final temperature.

make P' the subject of the equation

P' = PT'/T............ Equation 2

From the question,

Given: P = 50 psia, T = 300°R = (300×5/9)K = 166.66 K, T' = 600°R = (600×5/9)K = 333.33 K

Substitute these values into equation 2

P' = (50×333.33)/166.66

P' ≈ 100 psia

P ≈ 100 psia

Answer: D. 38x + 1,450 > 4,000

Step-by-step explanation:

It has to be greater than 4,000 so A makes no sense

The parentheses are in the wrong place completely changing the meaning for B

C disregards the info we have about how she's already typed 1,450 words

The answer has to be D

Answer:

[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}][/tex]

when:

[tex]|x|<\frac{1}{2}[/tex]

Step-by-step explanation:

In order to solve this problem, we must begin by splitting the function into its partial fractions, so we must first factor the denominator.

[tex]\frac{x+6}{2x^2-9x+5}=\frac{x+6}{(2x+1)(x-5)}[/tex]

Next, we can build our partial fractions, like this:

[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]

we can then add the two fraction on the right to get:

[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A(x-5)+B(2x+1)}{(2x+1)(x-5)}[/tex]

Since we need this equation to be equivalent, we can get rid of the denominators and set the numerators equal to each other, so we get:

[tex]x+6=A(x-5)+B(2x+1)[/tex]

and expand:

[tex]x+6=Ax-5A+2Bx+B[/tex]

we can now group the terms so we get:

[tex]x+6=Ax+2Bx-5A+B[/tex]

[tex]x+6=(Ax+2Bx)+(-5A+B)[/tex]

and factor:

[tex]x+6=(A+2B)x+(-5A+B)[/tex]

so we can now build a system of equations:

A+2B=1

-5A+B=6

and solve simultaneously, this one can be solved by substitution, so we get>

A=1-2B

-5(1-2B)+B=6

-5+10B+B=6

11B=11

B=1

A=1-2(1)

A=-1

So we can use these values to build our partial fractions:

[tex]\frac{x+6}{(2x+1)(x-5)}=\frac{A}{2x+1}+\frac{B}{x-5}[/tex]

[tex]\frac{x+6}{(2x+1)(x-5)}=-\frac{1}{2x+1}+\frac{1}{x-5}[/tex]

and we can now use the partial fractions to build our series. Let's start with the first fraction:

[tex]-\frac{1}{2x+1}[/tex]

We can rewrite this fraction as:

[tex]-\frac{1}{1-(-2x)}[/tex]

We can now use the following rule to build our power fraction:

[tex]\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}[/tex]

when |r|<1

in this case a=1 and r=-2x so:

[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2x)^n[/tex]

or

[tex]-\frac{1}{1-(-2x)}=-\sum_{n=0}^{\infty} (-2)^{n} x^{n}[/tex]

for: |-2x|<1

or: [tex] |x|<\frac{1}{2} [/tex]

Next, we can work with the second fraction:

[tex]\frac{1}{x-5}[/tex]

On which we can factor a -5 out so we get:

[tex]-\frac{1}{5(1-\frac{x}{5})}[/tex]

In this case: a=-1/5 and r=x/5

so our series will look like this:

[tex]-\frac{1}{5(1-\frac{x}{5})}=-\frac{1}{5}\sum_{n=0}^{\infty} (\frac{x}{5})^n[/tex]

Which can be simplified to:

[tex]-\frac{1}{5(1-\frac{x}{5})}=-\sum_{n=0}^{\infty} \frac{x^n}{5^(n+1)}[/tex]

when:

[tex]|\frac{x}{5}|<1[/tex]

or

|x|<5

So we can now put all the series together to get:

[tex]\frac{x+6}{2x^2-9x+5}=-\sum_{n=0}^{\infty} [(-2)^{n}x^{n} + \frac{x^{n}}{5^{n+1}}}[/tex]

when:

[tex]|x|<\frac{1}{2}[/tex]

We use the smallest interval of convergence for x since that's the one the whole series will be defined for.

Answer:

12

Step-by-step explanation:

The equation of the hyperbola is,

(x/12)² - 4y²/(527) = 1

The standard equation of the hyperbola is

(x/a)² - (y/b)² = 1

Here (a, 0) and (-a, 0) are vertices and asymptotes y = ± √(b/a)x

Foci are (c, 0) & (-c, 0)

Then a² + b² = c²

Here we have to give that.,

2a = 24

a = 12

And 2c = 7

c = 7/2

Therefore a = 12 and c = 3.5

Substituting a and c in Pythagorean identity;

b² = 527/4

Then, the equation of the hyperbola is

(x/12)² - 4y²/(527) = 1

For further information regarding hyperbolas, kindly refer

brainly.com/question/28989785

#SPJ4

We have b = 0, which implies that the foci coincide with the vertices, making the hyperbola a degenerate case. In this scenario, the equation of the border would be a vertical line passing through the vertices/foci, given by the equation x = ±a.

To find the equation of the hyperbolic border created by the shadow on the wall, we can start by understanding the properties of a hyperbola. A hyperbola is defined as the set of all points such that the difference of the distances from any point on the hyperbola to two fixed points, called the foci, is constant.

Let's label the vertices of the hyperbola as A and B, and the foci as F1 and F2. The distance between the vertices is given as 2a inches, and the foci are located at a distance c inches from the vertices.

Using the given information, we can find the value of a and c. Since the center of the hyperbola is at the origin, the coordinates of the vertices are (±a, 0), and the coordinates of the foci are (±c, 0).

The distance between the foci is given by the equation:

c = √(a^2 + b^2)

We know that the distance between the foci is given as 2c inches, so:

2c = 2√(a^2 + b^2)

Since c is given as a distance from the vertices, we can substitute c = a - b to simplify the equation:

2(a - b) = 2√(a^2 + b^2)

Squaring both sides to eliminate the square root:

4(a - b)^2 = 4(a^2 + b^2)

Expanding the equation:

4(a^2 - 2ab + b^2) = 4a^2 + 4b^2

Simplifying the equation:

4a^2 - 8ab + 4b^2 = 4a^2 + 4b^2

Canceling out the common terms:

-8ab = 0

Dividing by -8:

ab = 0

This implies that either a = 0 or b = 0. However, since a represents the distance between the vertices and b represents the distance between the foci and vertices, we can rule out a = 0 as it would result in a degenerate hyperbola.

for such more question on hyperbola

https://brainly.com/question/16454195

#SPJ8

Answer:

Answer is 6.

Step-by-step explanation:

The product is

[tex]15\times \frac{2}{5}[/tex]

Now, it does not means that the product of two quantities is always more than the individual quantities.

here, 2/5 is a part of whole.

So,

The product is

[tex]15\times \frac{2}{5}\\\\=3\times 2\\\\= 6[/tex]

The answer is 6 which is less than 15.

Here, it is the 2/5 part of whole 15.

Answer:

[tex]{ \tt{rate \: of \: change \: in \: A = 9}}[/tex]

Rate of change in function A is two times than that in function B

Answer:

? = 130

Step-by-step explanation:

I'm letting ? be x

Since the triangles are similar, larger outer and smaller inner, then the ratios of corresponding sides are equal.

If x is the length of side of larger then x - 70 is corresponding length of smaller.

Then

[tex]\frac{x}{x-70}[/tex] = [tex]\frac{78}{36}[/tex] ( cross- multiply )

78(x - 70) = 36x ← distribute left side

78x - 5460 = 36x ( subtract 36x from both sides )

42x - 5460 = 0 ( add 5460 to both sides )

42x = 5460 ( divide both sides by 42 )

x = 130

Answer:

≤ y ≤ 70 and 0 ≤ x ≤ 500

Step-by-step explanation:

In this relation we have two things to analyze, the number of gallons of water in the heater, that is 500 gallons, and the time that it took to empty the heater.

Let's count the time.

First, there are 20 minutes in wich 100 gallons are drained.

then, another drain valve is opened, so in 20 minutes they drain 200 gallons of water.

now, the wait for 10 minutes.

Now there are 200 gallons remaining, so the workers must wait for the other 20 minutes to drain the 200 gallons remaining.

The total amount of time is 70 minutes.

So if we have a relationship of water in the heater vs time, where X is the water remaining and Y is the time, the correct domains are:

Y from 0 minutes to 70 minutes

X from 0 gallons to 500 gallons  

So the correct options are C and E.

0 ≤ y ≤ 70 and 0 ≤ x ≤ 500

Answer:

Substitute into the quadratic formula

-6 ± √32 / 2

= -3 ± √16

= 1 and -1 are the answer

Answer:

x = - 3 ± 2[tex]\sqrt{2}[/tex]

Step-by-step explanation:

Given

x² + 6x + 1 = 0 ( subtract 1 from both sides )

x² + 6x = - 1

Using the method of completing the square

add/ subtract ( half the coefficient of the x- term)² to both sides

x² + 2(3)x + 9 = - 1 + 9

(x + 3)² = 8 ( take the square root of both sides )

x + 3 = ± [tex]\sqrt{8}[/tex] = ± 2[tex]\sqrt{2}[/tex] ( subtract 3 from both sides )

x = - 3 ± 2[tex]\sqrt{2}[/tex]

Then

x = - 3 - 2[tex]\sqrt{2}[/tex] , x = - 3 + 2[tex]\sqrt{2}[/tex]

Answer:

The volume is decreasing at a rate of about 118.8 cubic feet per minute.

Step-by-step explanation:

Recall that the volume of a cylinder is given by:

[tex]\displaystyle V=\pi r^2h[/tex]

Take the derivative of the equation with respect to t. V, r, and h are all functions of t:

[tex]\displaystyle \frac{dV}{dt}=\pi\frac{d}{dt}\left[r^2h\right][/tex]

Use the product rule and implicitly differentiate. Hence:

[tex]\displaystyle \frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)[/tex]

We want to find the rate at which the volume of the cylinder is changing when the radius if 4 feet and the volume is 32 cubic feet given that the radius is growing at a rate of 2ft/min and the height is shrinking at a rate of 3ft/min.

In other words, we want to find dV/dt when r = 4, V = 32, dr/dt = 2, and dh/dt = -3.

Since V = 32 and r = 4, solve for the height:

[tex]\displaystyle \begin{aligned} V&=\pi r^2h \\32&=\pi(4)^2h\\32&=16\pi h \\h&=\frac{2}{\pi}\end{aligned}[/tex]

Substitute:

[tex]\displaystyle\begin{aligned} \frac{dV}{dt}&=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)\\ \\ &=\pi\left(2(4)\left(\frac{2}{\pi}\right)\left(2\right)+(4)^2\left(-3\right)\right)\\\\&=\pi\left(\frac{32}{\pi}-48\right)\\&=32-48\pi\approx -118.80\frac{\text{ ft}^3}{\text{min}}\end{aligned}[/tex]

Therefore, the volume is decreasing at a rate of about 118.8 cubic feet per minute.

Answer:

Area of the given regular pentagon is 61.5 cm².

Step-by-step explanation:

Area of a regular polygon is given by,

Area = [tex]\frac{1}{2}aP[/tex]

Here, a = Apothem of the polygon

P = Perimeter of the polygon

Apothem of the regular pentagon given as 4.1 cm.

Side of the pentagon = 6 cm

Perimeter of the pentagon = 5(6)

                                             = 30 cm

Substituting these values in the formula,

Area = [tex]\frac{1}{2}(4.1)(30)[/tex]

        = 61.5 cm²

Therefore, area of the given regular pentagon is 61.5 cm².

Hi there!

[tex]\large\boxed{(-\infty, 9)}[/tex]

We can find the range using completing the square:

y = -x² - 2x + 8

Factor out a -1:

y = -(x² + 2x) + 8

Use the first two terms. Take the second term's coefficient, divide by 2, and square:

y = -(x² + 2x + 1) + 8  

Remember to add by 1 because we cannot randomly add an additional number into the equation:

y = -(x² + 2x + 1) + 8 + 1

Simplify:

y = -(x + 1)² + 9

Since the graph opens downward (negative coefficient), the range is (-∞, 9)

Answer:the answer is 9

Step-by-step explanation:

Hi there!

[tex]\large\boxed{12/5}}[/tex]

tan (angle) = Opposite side / Adjacent side, so:

Tan (A) = opposite side / adjacent side

= 24 / 10

Simplify:

= 12 / 5

Answer:

(a) [tex]PQ \sim ST[/tex]

Step-by-step explanation:

Given

See attachment

Required

Which must be true

[tex]\triangle PQR \simeq \triangle STU[/tex] implies that:

The following sides are corresponding

[tex]PQ \sim ST[/tex]

[tex]PR \sim SU[/tex]

[tex]QR \sim TU[/tex]

The following angles are corresponding

[tex]\angle P \sim \angle S[/tex]

[tex]\angle Q \sim \angle T[/tex]

[tex]\angle R \sim \angle U[/tex]

From the given options, only option (a) is true because:

[tex]PQ \sim ST[/tex]

Answer:

(in the image)

Step-by-step explanation:

I'm not sure I understood your question completely but I hope this helps.