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3 years ago

HELPPPP this is 7TH grade math and I don’t understand

Mathematics

2 answers:

Kipish [7]3 years ago

8 0

Part A

x = some unknown number

x+8 = add on 8 to the number

3(x+8) = triple the previous result

Therefore, one equation to set up is 3(x+8) = 66

This is the same as 3x+24 = 66 because 3(x+8) = 3x+3*8 = 3x+24. Use the distributive property to multiply that outer 3 to each term inside.

<h3>Answers:</h3>

C) 3x+24 = 66

D) 3(x+8) = 66

============================================================

Part B

Let's solve for x. I'll show the steps for each equation mentioned. When you write out the steps, you probably only need to pick one equation to solve.

3(x+8) = 66

x+8 = 66/3

x+8 = 22

x = 22-8

x = 14

3x+24 = 66

3x = 66-24

3x = 42

x = 42/3

x = 14

Either way, we get the same x value. To confirm this answer, replace x with 14 in either equation and simplify fully. You should get 66 on both sides.

<h3>Answer: x = 14</h3>

hodyreva [135]3 years ago

7 0

Answer:

B.3x+8=66

Step-by-step explanation:

i think ito po Ang answer hehehe

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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.