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AlekseyPX
2 years ago
11

Which inequalities are true? Select the four correct answers.

Mathematics
2 answers:
In-s [12.5K]2 years ago
4 0

Answer:

1 2 3 5

Step-by-step explanation:

Julli [10]2 years ago
3 0

Answer:

A.

B.

C.

F.

Step-by-step explanation:

It might be wrong but im about to finish

Edit: This is incorrect

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H = ac + mn solve for n
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Answer:

n =  \frac{h - ac}{m}

Step-by-step explanation:

h = ac + mn

mn = h - ac

n =  \frac{h - ac}{m}

6 0
10 months ago
Jace is deciding between two landscaping companies for his place of business. Company A charges $50 per hour and a $250 equipmen
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Step-by-step explanation:

4 0
3 years ago
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Y=-x-5 graph this equation
sergejj [24]
Negative slope and shifted 5 down on the y axis with a root at x=-5
5 0
3 years ago
Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x
sergij07 [2.7K]

By the definitions of cosecant, secant, and cotangent, we have

\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}

Then we rewrite the fraction as

\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}

The reason for this is that we want to apply the well-known limit,

\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1

for a\neq0. So when we take the limit, we have

\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}

=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3

8 0
3 years ago
PLEASE PLEASE HELP MEH
wel
Answer is in bold and i will add step
Since they are vertical angles they will be the same. Since 6=146 2=146

4 0
3 years ago
Read 2 more answers
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