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White raven [17]
2 years ago
13

I ONLY GOT 1 MIN!!!! PER QUESTION HELP

Mathematics
2 answers:
Solnce55 [7]2 years ago
7 0

Answer:

Your answer is <u>trapazoid</u>.

Step-by-step explanation:

This is the only one that is being described.

hope it helps

Pls mark brainliest :)

`

`

`

<em>Tori </em>

belka [17]2 years ago
4 0

Answer:

your answer is

<h2>a traepoizd </h2>

Step-by-step explanation:

<em><u>reason</u></em><em><u> </u></em>----

1 <em><u>parallelogram</u></em><em><u> </u></em>

  • it is opposite side quadrilateral
  • opposite angle

2 <em>rhombus</em><em> </em>

  • four congruence side in rhombus

3 <em><u>traepoizd</u></em><em><u> </u></em>

  • one pair of congruent sides that were not parallel. It also had one pair of parallel lines.

4 <em><u>kite</u></em><em><u> </u></em>

  • two pair of quadrilateral adjacent

<h2>hope it's helpful thanks please mark me as brainlist please </h2>
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Answer:

C

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

E. y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x hence y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x. Then y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x then y is not a solution.

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Alexus [3.1K]

Answer:

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Step-by-step explanation:

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