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3 years ago

Help with this math problem

Mathematics

2 answers:

belka [17]3 years ago

8 0

below -2 ll between -2 and 1 ll above 1

-2.5 ll -0.5 ll 1.5

-4.5 ll 0.5 ll 4

aniked [119]3 years ago

6 0

-0.5 goes in second box

1.5 goes in third box

-2.5 goes in first box

-4.5 goes in first box

0.5 goes in second box

4 goes in third box

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How do you write expression 5/6 using a fraction

Luden [163]

Answer:

5/6

Step-by-step explanation:

it is already written as a fraction.

do you mean as a decimal???

6 0

3 years ago

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:

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

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8 months ago

VXYZ is a kite. What is the length of XY?

olga nikolaevna [1]

Answer:

$5x - 6 = 3x + 4 \\ 5x - 3x = 4 + 6 \\ 2x = 10 \\ x = 10 \div2 \\ x = 5 \\ length \: of \: xy = 3 \times 5 + 4 \\ = 15 + 4 \\ = 19$