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zlopas [31]
2 years ago
9

What is n -8+4(1+5n)=-6n-14

Mathematics
1 answer:
jeka942 years ago
4 0

Answer:

n = (-5)/13

Step-by-step explanation:

Solve for n:

4 (5 n + 1) - 8 = -6 n - 14

4 (5 n + 1) = 20 n + 4:

20 n + 4 - 8 = -6 n - 14

Add like terms. 4 - 8 = -4:

20 n - 4 = -6 n - 14

Add 6 n to both sides:

20 n + 6 n - 4 = (6 n - 6 n) - 14

6 n - 6 n = 0:

20 n + 6 n - 4 = -14

20 n + 6 n = 26 n:

26 n - 4 = -14

Add 4 to both sides:

26 n + (4 - 4) = 4 - 14

4 - 4 = 0:

26 n = 4 - 14

4 - 14 = -10:

26 n = -10

Divide both sides of 26 n = -10 by 26:

(26 n)/26 = (-10)/26

26/26 = 1:

n = (-10)/26

The gcd of -10 and 26 is 2, so (-10)/26 = (2 (-5))/(2×13) = 2/2×(-5)/13 = (-5)/13:

Answer: n = (-5)/13

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Since - 5 is less than -4, hence the point (0, 0) is a solution to the inequality y-5<3x-4

<h3>Solution to inequality expression</h3>

Inequality are equations not separated by an equal sign. Given the inequality

y-5<3x-4

Check if (0, 0) is a solution

0 - 5 < 3(0) - 4

-5 < -4

Since - 5 is less than -4, hence the point (0, 0) is a solution to the inequality y-5<3x-4

learn more on inequality here: brainly.com/question/24372553

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A comet is about 7.63 × 103 miles from Earth. Write this number in standard form.
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Answer:

7630

Step-by-step explanation:

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This number is given in scientific notation.

We want to write this number in standard form.

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Therefore in standard form:

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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}

a)

\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

b)

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:

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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

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10 months ago
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masha68 [24]

Answer:d

Step-by-step explanation:

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