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3 years ago

5

Please help with geometry!! Thank you so much <3

1 answer:

Luda [366]3 years ago

6 0

Since the two triangles are proportional, there is a scale for going from one triangle to the other

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PLEASE HELP see picture<br><br><br> 66 pts<br> LOGARITHMIC

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2 years ago

Which statement belongs in the Perimeter section of the Venn Diagram?

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3 years ago

Read 2 more answers

Combine the like terms to create an equivalent expression<br> −5j+(−2j)+3

Oduvanchick [21]

Answer: -7j+3

Step-by-step explanation:Simplify

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2 years ago

A line has a slope 2/3 and a y- intercept -2 witch answer is the equation of the line

Mice21 [21]

Answer:

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Step-by-step explanation:

I'm pretty sure this is asking about the slope intercept form, since you didn't give us the actual answers. If not, comment below. Slope intercept form follows the y = mx + b form, where m is the slope and b is the y-intercept. By plugging your givens in, you get y = 2/3x - 2

6 0

2 years ago

Evaluate using L'Hopital's rule:<br><br> <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%201%5E%7B%2B%7D%7D%20%28x%20-%201

SSSSS [86.1K]

Answer:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

Step-by-step explanation:

We are given:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)$

And we want to evaluate it using L'Hopital's Rule.

First, using direct substitution, we will acquire:

$=(1-1)^\ln(1)}=0^0$

Which is indeterminate.

In order to apply L'Hopital's Rule, we first need to manipulate the expression. We will let:

$y=(x-1)^\ln(x)$

By taking the natural log of both sides:

$\ln(y)=\ln(x)\ln(x-1)$

And by taking the limit as x approaches 1 from the right of both functions:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\lim_{x\to 1^+}\ln(x)\ln(x-1)$

Rewrite:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{\ln(x-1)}{\ln(x)^{-1}}$

Using direct substitution on the right will result in 0/0. Hence, we can now apply L'Hopital's Rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}\frac{1/(x-1)}{-\ln(x)^{-2}(1/x)}\Big(\frac{-x\ln(x)^2}{-x\ln(x)^2}\Big)$

Simplify:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-\frac{x\ln(x)^2}{x-1}$

Now, by using direct substitution, we will acquire:

$\displaystyle \Rightarrow -\frac{1\ln(1)^2}{1-1}=\frac{0}{0}$

Hence, we will apply L'Hopital's Rule once more. Utilize the product rule:

$\displaystyle \lim_{x\to 1^+}\ln(y)= \lim_{x\to 1^+}-(\ln(x)^2+2x\ln(x))$

Finally, direct substitution yields:

$\Rightarrow -(\ln(1)^2+2(1)\ln(1))=-(0+0)=0$

Thus:

$\displaystyle \lim_{x\to 1^+}\ln(y)=0$

By the Composite Function Property for limits:

$\displaystyle \lim_{x\to 1^+}\ln(y)=\ln( \lim_{x\to 1^+}y)=0$

Raising both sides to e produces:

$\displaystyle e^{\ln \lim_{x\to 1^+}y}=e^0$

Therefore:

$\displaystyle \lim_{x\to 1^+}y=1$

Substitution:

$\displaystyle \lim_{x\to 1^+}(x-1)^\ln(x)}=1$

4 0

2 years ago