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2 years ago

What is the simplest form of

Mathematics

2 answers:

azamat2 years ago

5 0

Answer:

Step-by-step explanation:

weqwewe [10]2 years ago

3 0

Answer:

Step-by-step explanation:

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Now find another real world example (it can be a similar range problem). Graph it on a number line, and then explain the minimum

Kaylis [27]

The real world example for the absolute value are to express the change in balance of account.

How to use absolute value in real world problem?

Absolute value in real world problem is used to express the change or variation from one point to another point. In the absolute value, the sign of the number is not consideredd.

Let a real world example-

Everyone is going with 20 km/h and a person x is going with 15 km/h.
The person x has most likely to get the token as the difference between his speed and public speed is 5 km/h.
The sign of the number is not considered, whether it is +5 or -5 it is the difference of 5 km/h.

The number line for this absolute value is plotted below.

The real world example for the absolute value are to express the change in balance of account.

Learn more about absolute value in real world problem here:

brainly.com/question/2235065

#SPJ1

8 0

2 years ago

Which equation represents a line which is parallel to the line y = -5x – 7?

Vsevolod [243]

Answer:y=7 is the equation of a line. The line has a y-value of 7 no matter what x-value is given.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help me solve this problem using PEMDAS

asambeis [7]

The answer is 205
20 x 5 1/4 is 105
add the 100 and u get 205

5 0

3 years ago

Read 2 more answers

Again, I’m too lazy to do this but if you could help, thanks.

arsen [322]

Hi! the width is 5, and the length is 3. this one is easy because just find factors of 15 and plug them in!

7 0

3 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

3 years ago