What is 9+10 hurry or i will l fail

If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

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

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

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

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

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

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago

Helpppppppp pleaseeeeee

olga nikolaevna [1]

Answer:

7a + b - 16c

Step-by-step explanation:

So we have

7a-3b+c-8c+4b-9c

When it comes to problem like these, you need to combine like terms.

There's only one 'a' number, so that one remains the same.

Next, we have -3b and 4b. (-3b+4b)= b

Finally, there's (c-8c-9c).

(-8c-9c)= -17c

Take that and add the positive 'c' to it.

(-17c+c)= -16c

Put it all together now to get:

7a + b - 16c

I hope that helped and helped you understand at least a little

:)

6 0

3 years ago

THIS IS A K12 QUESTION WILL GIVE BRAINLIEST NEED ANSWER NOW!!!! For which distributions is the median the best measure of the ce

vampirchik [111]

Answer:

B

Step-by-step explanation:

A, C, and D can be a possibility but they are not exactly at the middle. B is the only one that shows a slightly triangular shape, which you can see the middle that shows the best measure of the center.

Hope this helped!

5 0

4 years ago

Read 2 more answers

3. IF ON = 5x - 7, LM = 4x +4, NM = x - 7, and OL = 3y – 4, find the values of x and y for which LMNO must

kkurt [141]

Answer:

x = 11 and $y = \frac{8}{3}$

Step-by-step explanation:

If LMNO is a parallelogram then it's opposite sides will be equal.

So, LM = ON

⇒ 4x + 4 = 5x - 7

⇒ x = 11

Now, MN = OL

⇒ x - 7 = 3y - 4

⇒ 11 - 7 = 3y - 4 {Since we already have, x = 11}

⇒ $y = \frac{8}{3}$ (Answer)

3 0

3 years ago

Help quick I have 30 minutes left

pashok25 [27]

Use tiger algebra it help

3 0

3 years ago

Read 2 more answers