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3 years ago

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a counter top is in the shape of a trapizoid . the lengths of the base is 70 1/2 and 65 1/2 inches long. the area of the counter

top is 1,224 square inches . write and solve an equation to find the height of the countertop.

1 answer:

Grace [21]3 years ago

4 0

I attached a photo of the work below :-)

Area of a trapezoid= [(a+b)÷2]h
1,224 in²= [(65.5 in+70.5in)÷2]h
1,224in²= (136÷2)h
1,224 in²= 68h
(1,224 in²÷68)= 68h÷68 (elimination)
h= 18 inches

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3 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

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Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

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$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$

We know that $6\sqrt{5+\sqrt{5}} = 16.13996\dots$ and $-\:6\sqrt{3-\sqrt{5}} = -5.24419\dots$ . Therefore,

Solution : $16.13996 - 5.24419i$

Which rounds to about option b.

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Step-by-step explanation:

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3 years ago

Which of the following best describes the

ehidna [41]

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$Two\ lines\ are\ perpendicular\ if\ product\ of\ the\ slopes\ is\ equal\ -1.\\\\k:y=\frac{3}{7}x+2\to the\ skolpe\ m_k=\frac{3}{7}\\\\l:y=\frac{7}{3}x-\frac{14}{3}\to the\ slope\ m_l=\frac{7}{3}\\\\m_k\times m_l=\frac{3}{7}\times\frac{7}{3}=1\neq-1\\\\conclusion:the\ lines\ are\ not\ perpendicular\\\\Two\ lines\ are\ parallel\ if\ the\ slopes\ are\ equal.\\\\m_k=\frac{3}{7};\ m_l=\frac{7}{3}\to m_k\neq m_l\\\\conclusion:the\ lines\ are\ not\ parallel$

$Answer:\boxed{C-The\ lines\ intersect\ but\ are\ not\ perpendicular.}$

3 0

3 years ago

Read 2 more answers