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3 years ago

Help me please number 7 (please show work)

Mathematics

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

m<H=30 degrees by the Triangle Sum Theorem, so <H=<V. <G=<U, they are right angles. If GH=UV, triangle FGH=triangle TUV by the ASA.

x=3 and y=8

Step-by-step explanation:

H equals 30 by 180-150=30 (getting 150 by adding 60 and 90)

I know this is not much of a answer, but I hope it helps a little.

http://www.somersetkey.com/ourpages/auto/2017/12/13/63309824/geo%20ch%205%20sec%204%20teach.pdf

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Answer:

Step-by-step explanation:

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4 years ago

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DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

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Answer:

Step-by-step explanation:

Given a triangle with angles 60° and 60°. Let the third angle be represented by x, so that;

x + 60° + 60° = $180^{o}$ (sum of angle in a triangle)

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x = $180^{o}$ - 120

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Thus, since the third angle of the triangle is 60°, then the triangle is an equilateral triangle. For an equilateral triangle, all sides are equal and all its angles are equal. So that the other sides of the triangle is 10 each.

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The required construction for the question is attached to this answer for more clarifications.

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Step-by-step explanation:

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