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sveticcg [70]
3 years ago
10

Does anyone know the passwords for the flvs 6th grade 04.08 module four test part one and two

Mathematics
2 answers:
lapo4ka [179]3 years ago
5 0

Answer:

I don’t know but you can try and use goog.le photos and take a picture and scan the mage and search the web. I garutntee you there is a least 1 answer key.

UkoKoshka [18]3 years ago
5 0

Answer:

these vary on teachers

Step-by-step explanation:

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Given f(x) and g(x) = f(x) + k, use the graph to determine the value of k.
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C.) 4

Step-by-step explanation:

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3 years ago
Read 2 more answers
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
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