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4 years ago

4( -x + 4 ) = 12 need a bit of help checking my work

2 answers:

3 0

4(-x + 4) = 12
-4x + 16 = 12
- 16. -16
-4x = -4
/-4. /-4

x = -1

That’s what I got. First get rid of parentheses, next subtract 16 on both sides, then divide by -4 to get x by itself and there’s your answer -1

galben [10]4 years ago

3 0

Remember to follow PEMDAS. Remember that the goal is to isolate the x.

First, distribute 4 to all terms within the parenthesis. Remember the signs.

4(-x + 4) = -4x + 16

-4x + 16 = 12

Next, isolate the x. Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS. First, subtract 16 from both sides.

-4x + 16 (-16) = 12 (-16)

-4x = 12 - 16

-4x = -4

Next, isolate the x. Divide -4 from both sides

(-4x)/-4 = (-4)/-4

x = -4/-4

x = 1

1 is your answer for x.

hope this helps

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Fantom [35]

Answer:

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫ $6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$ ]

So,

∫ $6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫ $t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫ $6x^5(x^6-2)\,dx$ = ∫ $6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.

7 0

3 years ago

I don't know how to solve this and the answer isn't 4 or 0 so what is it?4=-4+a

nevsk [136]

The answer is 8. This is because when you substitute 8 with “a” you then solve the problem to see if it is equal to the answer. -4+8=4... -4+4=0 so that is wrong and -4+O=-4 which is also incorrect.

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4 years ago