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2 years ago

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:

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

notka56 [123]

Answer:

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

Step-by-step explanation:

$h = ac + mn$

$mn = h - ac$

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

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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

Stells [14]

Answer:ttyftfhef

Step-by-step explanation:

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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

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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