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expeople1 [14]
3 years ago
8

Through (2,3) and (3,5)

Mathematics
1 answer:
Zinaida [17]3 years ago
7 0
What is the question your trying to ask...?
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Algebra 2 will give points and brainly
aivan3 [116]

Answer:

Step-by-step explanation:

In the 3rd question, we are given the equation x ^ 3 + x ^ 2 + 2x + 24. One of the factors is x + 3. Now, we can use long division to find that the equation we have left is x^2 - 2x - 8. We can just factor this to get (x - 4) (x + 2). In the 3rd question, possible factors for the coefficient are 1, 2, -1, -2. Possible factors for the constant are 1, 7, -1, -7. Now, we can try out all of them. The possible factors are 1, 7, -1, -7, 2, 14, -2, -14.

7 0
2 years ago
Are the equations 3x=−9 and 4x=−12 equivalent? Explain.<br> yes they are
My name is Ann [436]

Step-by-step explanation:

Yes they are equivalent. You get -3 when you try solving the actual equation.

5 0
2 years ago
Which of the following expressions is this one equivalent to?
padilas [110]

Hello from MrBillDoesMath!

Answer:

Choice D,  x-2


Discussion:

Observe the the highest order term in the numerator is  x^4 and the highest order term in the denominator is x^3. So the highest order term in their quotient is x^4/x^3 = x.  Choice D is the only possible answer as all other choices start with x^2

Regards,  

MrB

6 0
3 years ago
What is the domain of the function in the graph?
katrin [286]

Answer:

B is the correct answer!

Step-by-step explanation:

It starts at 6 and ends at 11.

Hope this helps & please mark brainiest....

5 0
2 years ago
Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
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