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2 years ago

13

rotate a square about one of its diagonals to form a conical figure. if one side of the square is 6 m, what is the surface area

of the solid?

1 answer:

Maurinko [17]2 years ago

5 0

6m, squares are symmetrical. All sides should equal the same on a square

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Find the exact value of cos theta, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.

vova2212 [387]

Answer:

$cos \theta = -\frac{8}{17}$

Step-by-step explanation:

For this case we know that:

$sin \theta = \frac{15}{17}$

And we want to find the value for $cos \theta$ , so then we can use the following basic identity:

$cos^2 \theta + sin^2 \theta =1$

And if we solve for $cos \theta$ we got:

$cos^2 \theta = 1- sin^2 \theta$

$cos \theta =\pm \sqrt{1-sin^2 \theta}$

And if we replace the value given we got:

$cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}$

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

$cos \theta = -\frac{8}{17}$

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2 years ago

HELP PLEASE 30 POINTS!!

Naya [18.7K]

Answer:

p - w = $25 OR w + $25 = p

Step-by-step explanation:

pencil = p

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p - w = $25 OR w + $25 = p

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2 years ago

Expand and reduce (2+√3)^2

nasty-shy [4]

Answer:

7 + 4 $\sqrt{3}$

Step-by-step explanation:

note that $\sqrt{a}$ × $\sqrt{a}$ = a

Given

(2 + $\sqrt{3}$ )² = (2 + $\sqrt{3}$ )(2 + $\sqrt{3}$ )

Each term in the second factor is multiplied by each term in the first factor, that is

2(2 + $\sqrt{3}$ ) + $\sqrt{3}$ (2 + $\sqrt{3}$ ) ← distribute both parenthesis

= 4 + 2 $\sqrt{3}$ + 2 $\sqrt{3}$ + 3 ← collect like terms

= 7 + 4 $\sqrt{3}$

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3 years ago

Please help the question is in the picture

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Answer: D: Parallel to the base

Step-by-step explanation:

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3 years ago

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Find the perimeter of the figure. 2 1/4 in, 4 1/12in, 3 2/3in

vitfil [10]

it would = out to be 9 18/73

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3 years ago