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3 years ago

HELP QUICK

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

3x + 6 = 2x

3x - 2x = -6

x = -6

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Solve for h, if the given equation is: s=2πr^2+2πrh solve

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First isolate the group with h in it
minus 2pir^2 from both sides
s-2pir^2=2pirh
divide both sides by 2pir
(s-2pir^2)/(2pir)=h

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3 years ago

What is the answer to this problem 7 6/7 + 7 1/9

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14.9682539683 Your welcome have a good day!

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Please Help! This is a trigonometry question.

liraira [26]

$\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}$

$\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}$

$\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}$

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$\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}$

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$\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}$

$\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}$

<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237

$\large\textsf{I hope it helps.}$

Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>

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3 years ago

Chris and Becky are comparing mortgages. The mortgage is for $183,500. If they choose 30 years at 5%, their monthly payment will

soldi70 [24.7K]

Answer:

For 30 years at 5% = $354,625.2

For 20 years at 4.5% = $278,618.4

Step-by-step explanation:

The total payback for each payment is given by the monthly payment multiplied by the total number of months for the payment period (years).

1) For 30 years at 5%

Total number of months in 30 years = 30 x 12 = 360 months

Payback = $985.07 x 360 = $354,625.2

2) For 20 years at 4.5%

Total number of months in 20 years = 20 x 12 = 240 months

Payback = $1160.91 x 240 = $278,618.4

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3 years ago

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A maybe? I don’t know honestly

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