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1 year ago

50 POINTS!!! PLEASE HELP DUE SOON

Mathematics

1 answer:

Mnenie [13.5K]1 year ago

7 0

Answer:

sorry I tried But Don't Know The Answer

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Find the area of the triangle with height (x+8) and base (2x+4). (A=1/2bh)

omeli [17]

Answer:

x^2+10x+16

Step-by-step explanation:

Plug in the values

h=x+8

b=2x+4

A=1/2(2x+4)(x+8)

A=(x+2)(x+8)

FOIL: x^2+8x+2x+16

A=x^2+10x+16

7 0

2 years ago

Dora earns an annual salary of $85,608. How much does she earn every month. A.$6,144. B. $7,134. C.$7,144. D. $7,234

Jet001 [13]

Hi let me help you out there
There are 12 months in a year
$85,608 is the amount she earns in a year
you do
85,608 divided by 12 since there are 12 months in a year
85,608 divided by 12= $7,134
The answer is B

hope this helps

3 0

3 years ago

Read 2 more answers

Jgvygdjvhjhg hgg bhg j gb gjbgjgjgvjbjh what is 2x+2

Serjik [45]

Answer: 2(x+1)

Step-by-step explanation: free points right?

3 0

3 years ago

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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>

Alla [95]

$\large \mathbb{PROBLEM:}$

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

$\large \mathbb{SOLUTION:}$

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

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

Please solve AND give a STEP BY STEP SOLUTION

RUDIKE [14]

2(-n - 3) -7(5+2n)
-2n - 6 - 35 - 14n
-16n - 41
B. -16 - 41

3 0

2 years ago