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2 years ago

Which is the simplified form of the expression3(7/5x+4)-2(3/2-5/4x)

Mathematics

1 answer:

kogti [31]2 years ago

5 0

Please tell me im smart, lol

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Distribute <br>(x-6)(x-7)<br><br>a) x^2-13x-15<br>b) x^2-13x+42<br>c) x^2+x+42<br>d) x^2+x-15

Sidana [21]

Answer:

B. x² - 13x + 42

Step-by-step explanation:

7 0

2 years ago

Solve the equation 8x-4=x+10

ahrayia [7]

Answer:

8x-4=x+10

8x-x = 10+4

7x = 14

x= 14/7

x= 2

<h2>HOPE IT HELPED U ...</h2>

8 0

3 years ago

Read 2 more answers

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

10 months ago

Guided Practice

KATRIN_1 [288]

Answer:

i would think its C

Step-by-step explanation:

the other options dont make sense lol

hope this is correct and helps :)

5 0

3 years ago

Read 2 more answers

I don't understand this.

sladkih [1.3K]

You mean the word or number

7 0

3 years ago