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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

Please help me thank you very much

frez [133]

Answer is A
Triangle UVW ~ Triangle UWT ~ Triangle WVT

5 0

3 years ago

Mitch has 50 pieces of licorice. He shares them equally with his sister.

leonid [27]

50/2=L is the a expression

7 0

3 years ago

Read 2 more answers

A ball is tossed up in the air at an initial rate of 50 ft/sec from 5 ft off the ground.

Fofino [41]

Answer with Step-by-step explanation:

Since we have given that

Initial velocity = 50 ft/sec = $v_0$

Initial height of ball = 5 feet = $h_0$

a. What type of function models the height (ℎ, in feet) of the ball after tt seconds?

As we know the function for height h with respect to time 't'.

$h(t)=-16t^2+v_0t+h_0\\\\h(t)=-16t^2+50t+5$

b. Explain what is happening to the height of the ball as it travels over a period of time (in tt seconds).

What function models the height, ℎ (in feet), of the ball over a period of time (in tt seconds)?

if it travels over a period of time then time becomes continuous interval . so it will use integration over a period of time

Our function becomes,

$h(t)=\int\limits^t_0 {-16t^2+50t+5} \, dt$

3 0

2 years ago

You invest $4000 in an account at 6.5% per year simple interest. How much will you have in the account at the beginning of the 9

aleksandrvk [35]

Answer:

$6340

Step-by-step explanation:

First figure out what 6.5% of 4000 is by multiplying not dividing because dividing will give you a way bigger number.

4000 x 6.5%=260

260 x 9=2340

2340+4000=6340

8 0

2 years ago