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Vlad1618 [11]
3 years ago
5

Please answer this.thank you​

Mathematics
2 answers:
Lady_Fox [76]3 years ago
8 0

Answer:

2(x + 4)

Step-by-step explanation:

This is because when you open up the brackets and multiply....

2 × x = 2x

and 2 × 4 = +8

So 2(x + 4) = 2x + 8

pentagon [3]3 years ago
6 0

Answer:

C

Step-by-step explanation:

cross multiply

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Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>​
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\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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In the figure, the measure of angle 9 is 80 degrees and the measure of angle 5 is 68. Find the measure of each angle.
murzikaleks [220]

Answer/Step-by-step explanation:

Given:

m<9 = 80°,

m<5 = 68°

Lines w and v are parallel to each other.

m<1 = m<9 (corresponding angles theorem)

m<1 = 80°

m<2 + m<1 = 180° (linear pair)

m<2 + 80° = 180° (substitution)

m<2 = 180° - 80°

m<2 = 100°

m<3 = m<1 (vertical angles are congruent)

m<3 = 80°

m<4 = m<2 (vertical angles are congruent)

m<4 = 100°

m<6 + m<5 = 180° (linear pair)

m<6 + 68° = 180° (substitution)

m<6 = 180° - 68°

m<6 = 112°

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m<7 = 68°

m<8 = m<6 (vertical angles)

m<8 = 112°

m<10 = m<2 (corresponding angles)

m<10 = 100°

m<11 = m<9 (vertical angles)

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m<12 = m<10 (vertical angles)

m<12 = 100°

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m<14 = 112°

m<15 = m<13 (vertical angles)

m<15 = 68°

m<16 = m<14 (vertical angles)

m<16 = 112°

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3 years ago
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