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3 years ago

Anyone ? what’s the volume of the cylinder

Mathematics

2 answers:

VikaD [51]3 years ago

8 0

What the guy above said is correct.

katen-ka-za [31]3 years ago

7 0

Answer:

V=πr2h

Step-by-step explanation:

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Three years ago, Sherman was twice as old as Clive. If Clive’s present age is c, which expression represents Sherman’s present a

Dennis_Churaev [7]

The answer is D

Sherman would still be double Clive's age (2c) and he would have aged 3 years (+3)

5 0

2 years ago

Please help thank you

Dvinal [7]

the tale-tell fellow is the number inside the parentheses.

if that number, the so-called "growth or decay factor", is less than 1, then is Decay, if it's more than 1, is Growth.

$\bf f(x)=0.001(1.77)^x\qquad \leftarrow \qquad \textit{1.77 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2(1.5)^{\frac{x}{2}}\qquad \leftarrow \qquad \textit{1.5 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=5(0.5)^{-x}\implies f(x)=5\left( \cfrac{05}{10} \right)^{-x}\implies f(x)=5\left( \cfrac{1}{2} \right)^{-x} \\\\\\ f(x)=5\left( \cfrac{2}{1} \right)^{x}\implies f(x)=5(2)^x\qquad \leftarrow \qquad \textit{Growth} \\\\[-0.35em] ~\dotfill$

$\bf f(t)=5e^{-t}\implies f(t)=5\left( \cfrac{e}{1} \right)^{-t}\implies f(t)=5\left( \cfrac{1}{e} \right)^t \\\\\\ \cfrac{1}{e}\qquad \leftarrow \qquad \textit{that's a fraction less than 1, Decay}$

now, let's take a peek at the second set.

$\bf f(x)=3(1.7)^{x-2}\qquad \leftarrow \qquad \begin{array}{llll} \textit{the x-2 is simply a horizontal shift}\\\\ \textit{1.7 is more than 1, Growth} \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3(1.7)^{-2x}\implies f(x)=3\left(\cfrac{17}{10}\right)^{-2x}\implies f(x)=3\left(\cfrac{10}{17}\right)^{2x} \\\\\\ \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill$

$\bf f(x)=3^5\left( \cfrac{1}{3} \right)^x\qquad \leftarrow \qquad \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3^5(2)^{-x}\implies f(x)=3^5\left( \cfrac{2}{1} \right)^{-x}\implies f(x)=3^5\left( \cfrac{1}{2} \right)^x \\\\\\ \textit{that fraction in the parentheses is less than 1, Decay}$

6 0

3 years ago

12 decreased by 8 times a number is 36

LekaFEV [45]

8x-12=36

Let's solve your equation step-by-step.

8x−12=36

Step 1: Add 12 to both sides.

8x−12+12=36+12

8x=48

Step 2: Divide both sides by 8.

8x/8=48/8

x=6

Answer: x=6

5 0

2 years ago

Factorize x^2+3x-ax-3a

Kobotan [32]

Answer:

=−ax+x2−3a+3x

Step-by-step explanation:

3 0

3 years ago

The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

puteri [66]

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Area of a Rectangle: A = lw

l is length
w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
[Rate] Multiply: $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Unit: Implicit Differentiation

Book: College Calculus 10e

8 0

3 years ago