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8_murik_8 [283]

2 years ago

14

This table represents the relationship between m, the amount of money an employee has in the morning, and f, the amount of money

an employee has after they bought food at work.
Which equation represents the relationship in the table?

f = m + 5.5

f=5.5−m

f=15.25m−5.5

f=m−5.5
Money in the morning (m) Money after buying lunch (f)
$15.25 $9.75
$10.50 $5.00
$7.00 $1.50
$21.75 $16.25

2 answers:

tatuchka [14]2 years ago

8 0

M-5.5 for sure I think so tell me if right

uranmaximum [27]2 years ago

7 0

Answer:

f= m - 5.5

Step-by-step explanation:

According to the table m is 15.25, so you would subtract 5.5 and you get 9.75 = f

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Find the measure of the exterior 21.<br> A. 15°<br> B. 100<br> C. 35<br> D. 145

FromTheMoon [43]

Answer:

The answer will be D. 145

Step-by-step explanation:

Because I said so

3 0

3 years ago

PLEASE HELP ASAP

Crank

Option D: $f(x)=-2(x+2)(x-4)$ is the function

Explanation:

Let the general form of quadratic equation be $y=a x^{2}+b x+c$

The function passes through the intercepts $(-2,0)$ and $(4,0)$ and also passes though the point $(-1,10)$

Substituting the points $(-2,0)$ , $(4,0)$ and $(-1,10)$ in the equation $y=a x^{2}+b x+c$ , we get,

$4a -2b+c=0$ -----------(1)

$16a +4b+c=0$ ----------(2)

$a-b+c=10$ -----------(3)

Subtracting (1) and (2), we get,

$\ \ 4a -2b+c=0\\ 16a +4b+c=0\\\---------\\ \ \ -12a-6b=0$ -----------(4)

Subtracting (2) and (3), we get,

$16a +4b+c=0\\\ \ a \ \ - \ \ b \ +c=10\\---------\\15a+5b=-10$ ------------(5)

Multiplying equation (4) by 5 and equation (5) by 4, to cancel the term b when adding, we get,

$-60a-30b=0\\\ 90a+30b=-60\\--------\\30a=-60\\\ \ a=-2$

Thus, the value of a is $a=-2$

Substituting $a=-2$ in equation (4), we get,

$\begin{aligned}-12 a+6 b &=0 \\-12(-2)-6 b &=0 \\ 24-6 b &=0 \\ 24 &=6 b \\ 4 &=b \end{aligned}$

Thus, the value of b is $b=4$

Now, substituting the value of a and b in equation (1), we have,

$\begin{aligned} 4 a-2 b+c &=0 \\ 4(-2)-2(4)+c &=0 \\-8-8+c &=0 \\ c &=16 \end{aligned}$

Thus, the value of c is $c=16$

Now, substituting the value of a,b and c in the general formula $y=a x^{2}+b x+c$ , we get,

$y=-2x^{2} +4x+16$

Taking out the common term as -2 we get,

$y=-2(x^{2} -2x-8)$

Factoring , we get,

$y=-2(x+2)(x-4)$

Thus, the function is $f(x)=-2(x+2)(x-4)$

7 0

3 years ago

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

harina [27]

Answer:

The volume of the solid is 714.887 units³

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: V = $\int\limits^a_b {2\pi xf(x)} \, dx$ ,

where 2πx · f(x) is the surface area of the cylinder shell and

dx is its thickness

* Lets solve the problem

∵ y = $x^{\frac{5}{2}}$

∵ The plane region is revolving about the y-axis

∵ y = 32 and x = 0

- Lets find the volume by the shell method

- The definite integral are x = 0 and the value of x when y = 32

- Lets find the value of x when y = 0

∵ $y = x^{\frac{5}{2}}$

∵ y = 32

∴ $32=x^{\frac{5}{2}}$

- We will use this rule to find x, if $x^{\frac{a}{b}}=c, then=== x=c^{\frac{b}{a}}$ , where c

is a constant

∴ $x=(32)^{\frac{2}{5}}=4$

∴ The definite integral are x = 0 , x = 4

- Now we will use the rule

∵ $V = \int\limits^a_b {2\pi}xf(x) \, dx$

∵ y = f(x) = x^(5/2) , a = 4 , b = 0

∴ $V=2\pi \int\limits^4_0 {x}.x^{\frac{5}{2}}\, dx$

- simplify x(x^5/2) by adding their power

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{7}{2}}} \, dx$

- The rule of integration of $x^{n} is ==== \frac{x^{n+1}}{(n+1)}$

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{9}{2}}} \, dx=2\pi[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}]$ from x = 0 to x = 4

∴ $V=2\pi[\frac{2}{9}x^{\frac{9}{2}}]$ from x = 0 to x = 4

- Substitute x = 4 and x = 0

∴ $V=2\pi[\frac{2}{9}(4)^{\frac{9}{2}}-\frac{2}{9}(0)^{\frac{9}{2}}}]=2\pi[\frac{1024}{9}-0]$

∴ $V=\frac{2048}{9}\pi=714.887$

* The volume of the solid is 714.887 units³

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3 years ago

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Step-by-step explanation:

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