2 years ago

Help please (inequalities)

Mathematics

2 answers:

Liula [17]2 years ago

7 0

The first one is the correct answer hope this helps have a nice day let me know if it works

zaharov [31]2 years ago

4 0

Answer:

The First one is the answer

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Graph and show work. Please. Helping a grandchild and I need to explain.

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3 years ago

2.432 2.09 2.190 2.37 least to greatest

Rus_ich [418]

2.09, 2.190, 2.37, 2.432

Hope this helps!

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3 years ago

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Lim x--> 0 (e^x(sinx)(tax))/x^2

Tom [10]

Make use of the known limit,

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

We have

$\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)$

since $\tan x=\dfrac{\sin x}{\cos x}$ , and the limit of a product is the same as the product of limits.

$\dfrac{e^x}{\cos x}$ is continuous at $x=0$ , and $\dfrac{e^0}{\cos 0}=1$ . The remaining limit is also 1, since

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so the overall limit is 1.

4 0

3 years ago

Plssssssssssssssssssssssssssssssssssssssssssssss

eduard

Answer:

34 hours

Step-by-step explanation:

Lets call M the number of hours that Maddie volunteered. Ryan volunteered 1 + 3×M hours, and altogether they volunteered 45 hours, so:

1 + 3×M + M = 45

1 + 4M = 45 Subtract 1 in both sides

4M = 44 Divide both sides by 4

M = 11 hours

So Maddie volunteered 11 hours, and Ryan volunteered

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2 years ago

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3 years ago

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