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Bad White [126]

3 years ago

12

The volume of a right rectangular prism is found by multiplying the length of the base by the width of the base by the height of

the prism. A right rectangular prism has a volume of 30 cubic inches. If the height of the prism is 6 inches, what is the area of the base of the prism?
A. 5 square inches
B. 24 square inches
C. 36 square inches
D. 180 square inches

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

5 in^2.

Step-by-step explanation:

Volume = area of the base * height

30 = area of the base * 6

area of the base = 30/6 = 5 in^2.

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If the perimeter of a square is 5 feet now many Inches long is each side of the square

gogolik [260]

Answer:

Well since we know that the perimeter of a square is four times the length of one of its sides. We just have to divide 5 by 4 to get the length of one side:

5feet /4 sides = 1.25 feet

And to finish off, we have to convert feet to inches:

1 foot = 12 inches

1.25 feet x inches

x inches = 12 inches x 1.25 feet ÷ 1 foot

x inches = 15 inches

Therefore, each side is 15 inches long.

Hope this helps!

Step-by-step explanation:

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2 years ago

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Order the set of numbers from least to greatest<br><br>3/5, 3/10, 2/7, 1/4

Mkey [24]

3/10 2/7 1/4 3/5 that is least to greatest

8 0

3 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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2 years ago

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elena-s [515]

Answer:

flaimgo/albert

Step-by-step explanation:

4 0

3 years ago

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A square napkin is folded in half on the diagonal and placed on the diameter of a round plate (see diagram below). If the folded

Crank

Answer:

$A=81(\pi-1)\ in^2$

Step-by-step explanation:

step 1

Find the area of the plate

The area of a circle is given by the formula

$A=\pi r^{2}$

we have

$r=18/2=9\ in$ ---> the radius is half the diameter

substitute

$A=\pi (9)^{2}\\A=81\pi\ in^2$

step 2

Find the area of the square napkin folded (is a half of the area of the square napkin)

we know that

The diagonal of the square is the same that the diameter of the plate

Applying Pythagorean theorem

$D^2=2b^2$

where

b is the length side of the square

we have

$D=18\ in$

substitute

$18^2=2b^2$

solve for b^2

$b^2=162\ in^2$ -----> is the area of the square

Divide by 2

$162/2=81\ in^2$

step 3

Find the area of the space on the plate that is NOT covered by the napkin

we know that

The area of the space on the plate that is NOT covered by the napkin, is equal to subtract the area of the square napkin folded (is a half of the area of the square napkin) from the area of the plate

so

$A=(81\pi-81)\ in^2$

simplify

$A=81(\pi-1)\ in^2$

8 0

3 years ago