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3 years ago

Answer quickly for 50 points, thank you.

Mathematics

2 answers:

Kaylis [27]3 years ago

6 0

The incorrect answer is B.

All you have to do is plug in x to the formulas and see if the are equal.

4x - 2 = 3x + 6

4(8) - 2 = 3(8) + 6

30 = 30

CORRECT

The second is not only not equal, but it is the wrong kind of formula for what you are trying to prove there.

Hope that helps!

Mashcka [7]3 years ago

4 0

Option B is incorrect this is because 3x+6 = 4x - 2 so add 2 to both sides and it becomes 3x+8=4x there for x is 8

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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

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

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

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

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

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

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A book is sold in Belize for 80.39 Belize dollars.
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