Pls help i dont understand :( will give Brainliest + 25 Points

Mathematics

1 answer:

icang [17]3 years ago

3 0

The answer I think is C

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If P is the incenter of JKL, find the measure of JLP

Snezhnost [94]

Answer:

<u>36°</u>

Step-by-step explanation:

> 1/2×180-(32+22)

> 1/2×180-(54)

> 90-54

> 36°

3 0

3 years ago

A town's employment statistics show that 855 of its residents were unemployed last month. This is a decrease of 25% from the pre

Leona [35]

Decrease of 25% means only 855 people? Okay, so you want to backtrack. If you divide by 25% you’re going to get what it was for a 75% decrease. So divide by .75 instead. 855/.75 = 1140 unemployed residents last month

6 0

2 years ago

Evaluate the limits<br><br>

julsineya [31]

$x > \ln(x)$ for all $x$ , so

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

Similarly, $\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}$

We can of course see the limits are identical by replacing $x\mapsto e^x$ , so that

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)$

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$