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1 year ago

Please answer!!!!!!!

Mathematics

2 answers:

tamaranim1 [39]1 year ago

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The file won't open sorry

jeka941 year ago

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The file won’t open but do you think you can repost it?

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Drag a statement or reason to each box to complete this proof.

ololo11 [35]

Answers:

1) -3(x + 8) = -21

2) Division Property of Equality

3) x + 8 = 7

4) Subtraction Property of Equality

Reasoning:

You are already given the equation, so that will be #1 in the Given.

#2 shows both sides being divided by -3, so you will put the Division Property of Equality.

#3 shows the Division Property after it is carried out, so you will put the equation after the division is done.

#4 shows you are subtracting 8 from both sides, so put the Subtraction Property of Equality.

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2 years ago

Y-y2= m(x-x) coordinates of a point on the line

mario62 [17]

Answer:yes 5,5

Step-by-step explanation:

6 0

2 years ago

What value for x will make the equation true?<br> -3 (7x - 5) = 87 + 3x

Gennadij [26K]

Answer:

x=-3

Step-by-step explanation:

follow the PEMDAS method then see if the 2 equation equal each other after plugging in the x.

7 0

2 years ago

Read 2 more answers

An unknown number y is 12 more than an unknown number x. The number y is also x less than 17. The equations to find x and y are

grandymaker [24]

Answer:

d because then you can solve for y then use your answer to find x

4 0

2 years ago

Read 2 more answers

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

1 year ago