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2 years ago

Explain why 3(x+4)=3(x-5) has no solution. Choose the best response below.

2 answers:

6 0

It's A.

You've got 3 as a coefficient on both sides, so you can divide both sides by 3 to cancel them out. Then you've got:

x + 4 = x - 5

You know that this can't have a solution because there is no value x where if you add 4 to it and subtract 5 from it, you'll get the same thing.

So, the problem with the equation is that the x values have to be the same, but the constant numbers added/subtracted are different.
(If you liked this answer please rate it as brainliest!! thank you!!)

Evgesh-ka [11]2 years ago

3 0

Its either A or C i know that

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Plz help question below

9966 [12]

X+y=500
215x+61y=187,500

3 0

2 years ago

A club raised 175% of its goal for a charity. The club raised $763. What was the goal?

IceJOKER [234]

Answer: The clubs goal is to raise $436.

Step-by-step explanation:

175% * x = 763 where x is the goal

1.75 *x = 763

1.75x = 763

x= 436

3 0

2 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

1 year ago

Someone help. Urgent! I need someone to answer this! Factor -5b – 5c!

tatyana61 [14]

Answer:

the answer you are looking for is-10

Step-by-step explanation:

6 0

1 year ago

Equation: w = 112 + 8t Which input results in an output of 176? What is the output in the ordered pair (4, 144)?

Grace [21]

9514 1404 393

Answer:

Step-by-step explanation:

Generally, the single variable on the left side of the equation is the "output" or dependent variable, while any other variables on the right side of the equation are considered to be independent variables, or "input."

So, we assume your first question is asking for the value of t that makes w=176. Put the values in the equation and solve for the unknown.

w = 112 +8t

176 = 112 + 8t

64 = 8t . . . . . . . subtract 112

8 = t . . . . . . . . . divide by 8

The input for an output of 176 is 8.

Ordered pairs are always written as ...

(input, output)

So, (4, 144) means the output is 144 for an input of 4.

7 0

2 years ago

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