Ask question

Ask question

All categories

4 years ago

13

Which equation is not equvilant to A=1/2 bh,

1 answer:

ollegr [7]4 years ago

3 0

Which equation is not equvilant It is c

You might be interested in

Express <br> 8.218.21<br> 8.21<br> 8, point, 21<br> as a mixed number.

djverab [1.8K]

8 21/100 is the answer

3 0

3 years ago

If you have to slow your car when your driving down a hill you should

ludmilkaskok [199]

Answer:

I would say so

Step-by-step explanation:

If you are speeding down a hill that could be dangerous for the car and other people around

5 0

4 years ago

a. If cosθ=−45 where θ is in quadrant 3, find sin2θ. b. If cosθ=2√2 where θ is in quadrant 1, find cos2θ. c. If sinθ=817 where θ

sesenic [268]

Answer:

Part A) $sin(2\theta)=\frac{24}{25}$

Part B) $cos(2\theta)=0$

Part C) $tan(2\theta)=-\frac{240}{161}$

Step-by-step explanation:

Part A) we have $cos(\theta)=-\frac{4}{5}$

θ is in quadrant 3 ----> the sine is negative

Find $sin(2\theta)$

we know that

$sin(2\theta)=2sin(\theta)cos(\theta)$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$(-\frac{4}{5})^{2}+sin^{2} (\theta)=1$

$(\frac{16}{25})+sin^{2} (\theta)=1$

$sin^{2} (\theta)=1-\frac{16}{25}$

$sin^{2} (\theta)=\frac{9}{25}$

$sin(\theta)=-\frac{3}{5}$ ---> remember that the sine is negative (3 quadrant)

Find $sin(2\theta)$

we have

$cos(\theta)=-\frac{4}{5}$

$sin(\theta)=-\frac{3}{5}$

$sin(2\theta)=2sin(\theta)cos(\theta)$

substitute

$sin(2\theta)=2(-\frac{3}{5})(-\frac{4}{5})$

$sin(2\theta)=\frac{24}{25}$

Part B) we have $cos(\theta)=\frac{\sqrt{2}}{2}$

θ is in quadrant 1

Find $cos(2\theta)$

we know that

$cos(2\theta)=2cos^{2} (\theta)-1$

substitute

$cos(2\theta)=2(\frac{\sqrt{2}}{2} )^{2}-1$

$cos(2\theta)=0$

Part C) we have $sin(\theta)=\frac{8}{17}$

θ is in quadrant 2 ----> the cosine is negative

Find $tan(2\theta)$

we know that

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$cos^{2} (\theta)+(\frac{8}{17})^{2}=1$

$cos^{2} (\theta)=1-\frac{64}{289}$

$cos^{2} (\theta)=\frac{225}{289}$

$cos(\theta)=-\frac{15}{17}$

Find $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

substitute

$tan(\theta)=\frac{(8/17)}{(-15/17)}$

$tan(\theta)=-\frac{8}{15}$

Find $tan(2\theta)$

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

substitute

$tan(2\theta)=\frac{2(-\frac{8}{15})}{1-(-\frac{8}{15})^{2}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-(\frac{64}{225})}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-\frac{64}{225}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{\frac{161}{225}}$

$tan(2\theta)=-\frac{240}{161}$

3 0

3 years ago

Find the term independent of x in the expansion of

laiz [17]

Step-by-step explanation:

$\frac{19683x {}^{18} }{512} - \frac{59049x {}^{15} }{512} + \frac{19683x {}^{12} }{128 } - \frac{15309x {}^{9} }{?128} + \frac{15309x {}^{6} }{256} - \frac{15309x {}^{9} }{128} + \frac{15309x {}^{6} }{?256} - \frac{5103x {}^{3} }{256} + \frac{567}{128} - \frac{81}{128x {}^{3} } + \frac{27}{512x {}^{6} } - \frac{1}{512x {}^{9} }$ it's long there's some calculations in de side hope its helpful

8 0

3 years ago

16.15 – 13<br> A. (413 – 32)2<br> B. 33(4.1 – 1)(45 + 1)<br> c. 3(45 – 1)2<br> d. 33(1682 – x)

sergiy2304 [10]

Answer:

neither its 3.15

Step-by-step explanation:

all you have to do is subtract the first number from the secondes

4 0

4 years ago