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3 years ago

14

can someone help me answer this quick. I have to submit it right now. Question- which equation correctly uses the segment additi

on postulate for this scenario??

2 answers:

Stels [109]3 years ago

8 0

The answer is d hope this helps

kobusy [5.1K]3 years ago

4 0

Answer:

Ok first off which do you think it is??

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shutvik [7]

the answer is 235 cubic meters

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The perimeter of the shape is <br><br> feet.

Archy [21]

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Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

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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3 years ago

Simplify: (x^-3/y) (x/y)^-1

Phantasy [73]

Answer:

x(5)/(6)y(1)/(2)

Step-by-step explanation:

Hoped I helped

leave a like or rate.

3 0

3 years ago

What is the approximate area, in square units, of the circle below

Dmitry [639]

Answer:

A) 50

Attached is the image of the figure.

Step-by-step explanation:

Area of a circle can be expressed as;

Area A = πr^2 ......1

Where;

r = radius

From the attached image, the shape is a Circle and the diameter of the circle is 8 units.

Diameter d = 8

Radius r = diameter/2 = 8/2 = 4 units

Substituting the radius into equation 1;

Area A = π × 4^2 = 16π

Area A = 50.265

Area A = 50 square units (approximately)

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3 years ago