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3 years ago

Find the slope of the line going through the points (-5, 2) and (-1,5).

Mathematics

2 answers:

ahrayia [7]3 years ago

8 0

7/4 you have to subtract 5-2/-1-(-5)

Karolina [17]3 years ago

4 0

The slope would be 3/4 because 5 - 2 is 3 and -1 - (-5) is 4. Hope this helps!

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a web music store offers two version of a popular song. the size of the standard version is 2.9 megabytes (MB). the size of the

miv72 [106K]

Answer: 260 downloads of the standard version were there.

Step-by-step explanation:

Let x represent the number of downloads of the standard version.

Let y represent the number of downloads of the high-quality version.

Yesterday, the high-quality version downloaded four times as often as the standard version. It means that

y = 4x

The size of the standard version is 2.9 megabytes (MB). the size of the high-quality version is 4.6 MB. the total size downloaded for the two versions was 5538 MB. It means that

2.9x + 4.6y = 5538- - - - - - - - - - - -1

Substituting y = 4x into equation 1, it becomes

2.9x + 4.6 × 4x = 5538

2.9x + 18.4x = 5538

21.3x = 5538

x = 5538/21.3

x = 260

6 0

4 years ago

NEED HELP FAST PLEASE

Bess [88]

The answer is 39 and I know because I took the test and got it right

3 0

3 years ago

Solve for the value of: 14/5 =z/25

ryzh [129]

Answer:

Solve for

by simplifying both sides of the equation, then isolating the variable.

Step-by-step explanation:

Hope I helped

8 0

3 years ago

Please, with work, find the prime factorization of 60 and 140?

Brrunno [24]

Prime Factors of 60: 2,3, and 5
Prime Factors of 140: 2,5 and 7
All mental math no work needed.

3 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

1 year ago