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3 years ago

What is the area of CDEF? Please could you also say how you got your answers, it will be much appreciated.

Mathematics

1 answer:

ziro4ka [17]3 years ago

7 0

This image has the step by step solution

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2. A large aquarium in the shape of a rectangular prism has the follo

timurjin [86]

Answer: 38568 3/4

Step-by-step explanation:

Volume(V) = lwh

Where l=length;w=width;h=height

V = 50 * 25 1/2 * 30 1/4

V = 38,568 3/4 or 38,568.75

7 0

3 years ago

Read 2 more answers

Find the magnitude of the vector from the origin to (8. - 6) and write the vector as the sum of unit vectors.

Kay [80]

The magnitude of the vector from the origin is 10

The unit vector in of the vector from the origin as (8/10, -6/10

Step-by-step explanation:

1.Finding the magnitude,

we have the formula,

magnitude=√a²+b²

we have the values as a=8 and b=-6

Finding the magnitude we get,

magnitude=√a²+b²

magnitude=√8²+6²

magnitude=√100

magnitude=10

The magnitude of the vector from the origin is 10

2.Finding the unit vector

Divide by the magnitude

Unit vector: (8/10, -6/10)

The unit vector in of the vector from the origin as (8/10, -6/10

3 0

3 years ago

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

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

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.