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What is the scale factor of two similar pyramids with volume of 13824 cubic feel and 216 cubic feet?

velikii [3]

Answer: The scale factor is 4

Step-by-step explanation:

We know that the pyramids are similar. The volume of one of these pyramids is 13,824 cubic feet and the volume of the other one is 216 cubic feet. Then:

$V_1=13,824ft^3\\V_2=216ft^3$

By Similar solids theorem, if two similar solids have a scale factor of $\frac{a}{b}$ , then corresponding volumes have a ratio of $\frac{a^3}{b^3}$

Then:

$\frac{V_1}{V_2}=\frac{a^3}{b^3}$

Knowing this, we can find the scale factor. This is:

$\frac{13,824}{216}=\frac{a^3}{b^3}\\\\\frac{13,824}{216}=(\frac{a}{b})^3\\\\\frac{a}{b}=\sqrt[3]{\frac{13,824}{216}}\\\\scale\ factor=\frac{a}{b}=4$

3 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:

Calculus

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:

Step 1: Define

Identify

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

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

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

Step 3: Integrate Pt. 2

[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

4 0

2 years ago

Solve for p. 0.6p +4.5 = 22.5

mash [69]

0.6p + 4.5 = 22.5
Subtract 4.5 from both sides, which will give you, 18.5.
Then, divide 0.6 from both sides.

8 0

3 years ago

Read 2 more answers

-\frac{9x}{4}=27

IRISSAK [1]

$\bf \cfrac{-9x}{4}=27\implies \stackrel{\textit{cross multiplying}}{-9x=4(27)}\implies x=\cfrac{4(27)}{-9}\implies x=4\cdot \cfrac{27}{-9} \\\\\\ x=4\cdot -3\implies x=-12$

4 0

2 years ago

5x + qy, when x=8,y=11

Nataliya [291]

Answer:

40+11q

Step-by-step explanation:

5x+qy

5(8)+q(11)=40+11q

4 0

3 years ago