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3 years ago

How do you answer $9 ; 10%

1 answer:

3 0

<span>You can calculate percentages by first finding what 10% of a value is and then what 5 or 1% is. From there you can multiply to find multiples and then add to get the required percentage.</span>

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What is the value of 3x squared when x = 2?

KonstantinChe [14]

Answer:

Step-by-step explanation:

$(3x)^{2}$ = $(3(2))^{2}$ =36

(OR)

3( $x^{2}$ )=3(4)=12

Its depend on your sentence structure, what u r asking!!

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3 years ago

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PLEASE HELP FAST! I will give brainlist!

Whitepunk [10]

Answer:

D or A

Step-by-step explanation:

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3 years ago

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The area of a square pond is 1000m2.A path of uniform width is surrounded outside the pond and its area is 369m2.find the outer

Romashka-Z-Leto [24]

Answer:

631 m²

Step-by-step explanation:

Outer length of park = Total area - Area of pond

Outer length of park = 1000 - 369

Outer length of park = 631 m²

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2 years ago

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3. IN YOUR OWN WORDS How can you find the area of a circle?

Ivahew [28]

Answer: you can find the area of a circle by multiplying the radius by pi and then squaring it by 2

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

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

7 0

2 years ago