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2 years ago

A number increased by 4 is no more than -18.

Mathematics

1 answer:

Katyanochek1 [597]2 years ago

7 0

Answer:

n+4≤-18

This might be it

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PLEASEEE HELPP WILL GIVE BRAINLIEST

Lady bird [3.3K]

Answer:

-1/2

Step-by-step explanation:

Negative because the slope is going down

3 0

2 years ago

1. Find the possible values for s in the inequality 12s – 20 ≤ 50 – 3s – 25.

Elan Coil [88]

12s - 29 < 50 - 3s - 25

Like terms

12s - 29 < 25 - 3s
+3s. +3s

S cancels

15s - 29 < 25
+ 29. +29

15s. < 54
/15. /15

S < 3.6

4 0

3 years ago

Read 2 more answers

Can i have help please

galina1969 [7]

Answer:

Step-by-step explanation:

8 0

2 years ago

PLS HELP!!!!What are the sine, cosine and tangent ratios of<br>angle W in the triangle below

Ksju [112]

Step-by-step explanation:

Remember SOH-CAH-TOA:

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

The side adjacent to W is 4. The side opposite of W is 3. The hypotenuse is 5.

Therefore:

Sine = 3 / 5

Cosine = 4 / 5

Tangent = 3 / 4

7 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:

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

4 0

7 months ago