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FrozenT [24]
2 years ago
6

How do you find a surface area of a rectangular prism?

Mathematics
1 answer:
n200080 [17]2 years ago
4 0
Use the equation
A= 2(wl +hl +hw)

Where
w is width
h is height
l is length
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3 years ago
4 x 7 =?
julia-pushkina [17]
The answer is a, i dont think i want the prize tho lol
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2 years ago
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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
How many solutions are ther to this equation.
Korolek [52]
There are multiple solutions and they will vary. 

Hope this helps!
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3 years ago
The difference between a number and seven is less than or equal to -3
Scorpion4ik [409]
X-7 < -3
Remember that the less than or equal to has a line underneath it!
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