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2 years ago

hi can i please have some help w/ these! i have my exam tomorrow!!! factoring algebra 2! worth 35 points!

Mathematics

1 answer:

prohojiy [21]2 years ago

4 0

Answer:

11. $4x^{3}-10x^{2}+6x$

Take out 2x ( greatest common factor )
$2x(2x^{2}-5x+3)$

Factor

$2x(2x^{2} -2x-3x+3)$

$2x(x-1)-3(x-1)=(2x-3)(x-1)$

Don't forget about the 2x we took out earlier

Solution to 11: $2x(2x-3)(x-1)$

12. $16x^{2} -25y^{2}$

This is the difference of squares, so let's use the formula...

$a^{2}-b^{2}=(a-b)(a+b)$

if, $a=16x^{2}$ and $b=25y^{2}$

then, the Solution to 12 is:

$(4x-5y)(4x+5y)$

13. $3x^3-24$

First of all, you must find the greatest common factor because there isn't anything else you can do.

$3(x^3-8)$

Now you can see the difference of cubes...

$a^3-b^3=(a-b)(a^2+ab+b^2)$

so, the Solution to 13 is:

$3(x^3-8)= 3(x-2)(x^2+2x+4)$

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Could anyone help me with this question? Thank you!

shutvik [7]

Answer:

Step-by-step explanation:

We want to simplify the rational expression:

$\displaystyle \frac{n^4-11n^2+30}{n^4-7n^2+10}$

First, we can factor both the numerator and the denominator. This yields:

$\displaystyle =\frac{(n^2-6)(n^2-5)}{(n^2-2)(n^2-5)}$

Cancel like terms:

$\displaystyle =\frac{(n^2-6)}{(n^2-2)}$

However, since we canceled a term from our expression, we must place restrictions in order to stay true to the original expression.

Rational expressions are undefined whenever the denominator equals 0. The denominator is (before canceling):

$(n^2-2)(n^2-5)=0$

Solve for n. By the zero product property:

$n^2-2=0\text{ and } n^2-5=0$

Therefore:

$n^2=2\text{ and } n^2=5$

By taking the square root of both sides (since we are taking an even root, it is required to have plus/minus):

$n=\pm\sqrt{2}\text{ and } n=\pm\sqrt5$

Therefore, our fully simplified expression is:

$\displaystyle \frac{n^2-6}{n^2-2}\, ,n\neq \pm\sqrt5, \pm\sqrt2$

Hence, our answer is C!

4 0

2 years ago

Solve graphically x+1 is grater than 2x-2

Greeley [361]

You have to solve using the graph as it says solve graphically

5 0

2 years ago

Find the area of the parallelogram shown.

ziro4ka [17]

Answer:

40 Yd^2

Step-by-step explanation:

Take out the 7 from that problem and multiply the regular shape so whats 8 times 5?

8 0

3 years ago

Read 2 more answers

What’s the answer for this one??

KIM [24]

Answer:

It is the second one.

Step-by-step explanation:

a and e are equal because they are mirrored. d and b are equal because they are on the same line and at the same point where the parallel lines go through. b and f are also mirrored. That leaves c and e which don't match.

6 0

2 years ago

Please help me with the below question.

Snezhnost [94]

6a. By the convolution theorem,

$L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}$

6b. Similarly,

$L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}$

7. Take the Laplace transform of both sides, noting that the integral is the convolution of $e^t$ and $f(t)$ .

$\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau$

$\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)$

where $g(t) = e^t$ . Then $G(s) = \frac1{s-1}$ , and

$F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}$

We have the partial fraction decomposition,

$\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)$

Then we can easily compute the inverse transform to solve for f(t) :

$F(s) = -\dfrac1s + \dfrac4{s+3}$

$\implies \boxed{f(t) = -1 + 4e^{-3t}}$

6 0

1 year ago