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3 years ago

What is 24% of 135??

Mathematics

2 answers:

Marina CMI [18]3 years ago

8 0

Answer:

32.4

Step-by-step explanation:

135 times 24 dived by 100

KiRa [710]3 years ago

6 0

Answer:

32.4

135 x 24% = 135 x 0.24 = 32.4

Step-by-step explanation:

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kherson [118]

Answer:

Step-by-step explanation:

it rises 2 and goes to the right once; 2/1 = 2

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4 years ago

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Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of the A^2?

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

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2 years ago

Find the value of w if the expression wx(3y² + 6y – 2) simplifies to the expression 6xy² + 12xy – 4x.

nadezda [96]

Answer:

w = 2

Step-by-step explanation:

Distribute the expression and compare like terms with the simplified version.

Given

wx(3y² + 6y - 2) ← distribute parenthesis

= 3wxy² + 6wxy - 2wx

Compare coefficients of like terms with

6xy² + 12xy - 4x

Compare xy² term, then

3w = 6 ( divide both sides by 3 )

w = 2

Compare xy term, then

6w = 12 ( divide both sides by 6 )

w = 2

Compare x term, then

- 2w = - 4 ( divide both sides by - 2 )

w = 2

Hence the required value of w is 2

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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