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3 years ago

What is the solution to this equation 10x - 3(x-6)- x + 30

Mathematics

1 answer:

Llana [10]3 years ago

8 0

Answer:

option B)

x = 2

Step-by-step explanation:

Given in the question the equation:

10x - 3(x- 6) = x + 30

Simplify the left side

10x - 3(x - 6) = x + 30

10x - 3x + 18 = x + 30

rearrange the terms and containing x to the left side of the equation and add accordingly

x(10 - 3 - 1) = (30 - 18)

x(10 - 4) = (12)

x(6) = 12

6x = 12

Divide each term by 6 and simplify

x = 12/6

x = 2

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Solve for x please and thank you

densk [106]

Since the angles sum of △ is 180°,here's the equation:
60°+65°+(8x+7)=180°
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8x=55-7
x=48/8
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Thus,the answer is x=6

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3 years ago

Find the volume of the sphere. Give the answer in terms of pi and rounded to the nearest cubic foot.

Inessa05 [86]

Hi there!

The formula for the volume of a sphere is V = 4/3(pi)(r^3). Using this formula, we can plug in what we know and solve.

WORK:
V = 4/3(pi)(10^3)
V = 4/3(1000pi)
V = 4000pi/3 or 1333.3 pi
V also = 4188.8

Hope this helps!! :)
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3 years ago

Read 2 more answers

If X = 13 units, Y = 15 units, and h = 11 units, what is the area of the parallelogram shown above?

Sedbober [7]

Answer:

the area of the parallelogram = base times the height

For the given parallelogram :

x = 8 units, y = 13 units, and h = 11 units

1) If x is the perpendicular to h

So, the base = 8 and the height = 11

So, the area = 8 * 11 = 88 square units

2) If y is the perpendicular to h

So, the base = 13 and the height = 11

So, the area = 13 * 11 = 143 square units

Step-by-step explanation:

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2 years ago

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

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

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I m. .

. Za v cap rf

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