All categories

2 years ago

Really need an answer to this please help.

Mathematics

1 answer:

Alex2 years ago

7 0

Answer:

Step-by-step explanation:

$\sqrt[4]{144a^{12}b^{3}} = \sqrt[4]{4^{2}*3^{2}a^{12}b^{3}}=\\= \sqrt[4]{2^{4}*3^{2}a^{12}b^{3}}=2a^{3}\sqrt[4]{3^{2}b^{3}} =\\}=2a^{3}\sqrt[4]{9b^{3}}$

You might be interested in

The angle measures of a triangle are (x + 20)°, (7x + 11)°, and (4x - 7)°. Which of the following is NOT one of the angle measur

bezimeni [28]

Answer:

Step-by-step explanation:

6 0

2 years ago

Please select the best answer from the choices provided

Ira Lisetskai [31]

<u>Answer:</u>

The correct answer option is c. $\frac{1}{9a^2}$ .

<u>Step-by-step explanation:</u>

We are given an expression $(27a^{-3})^{-\frac{2}{3} }$ and we are supposed to simplify it.

We can also write $(27a^{-3})^{-\frac{2}{3} }$ as $(\frac{3^3}{a^3} )^{-\frac{2}{3} }$ .

We know the power rule (a^m)^n=a^{mn} which means that to raise a power to a power you need to multiply the exponents.

So multiplying the exponents to get:

$\frac{3^{(3.-\frac{2}{3})} }{a^{(3.\frac{2}{3})} }\\\\\frac{3^{-2}}{a^2}\\ \\\frac{1}{3^2a^2} \\\\\frac{1}{9a^2}$

Therefore, the correct answer option is c. $\frac{1}{9a^2}$ .

4 0

3 years ago

What is the solution of the system?

Kryger [21]

Greetings!

Solve the System, using Elimination:
$\left \{ {{4x+2y=18} \atop {2x+3y=15}} \right.$

Multiply Equation #2 by 2:
$\left \{ {{4x+2y=18} \atop {2(2x+3y)=2(15)}} \right.$

$\left \{ {{4x+2y=18} \atop {4x+6y=30}} \right.$

Eliminate variable x:
$-\frac{ \left \{ {{4x+2y=18} \atop {4x+6y=30}} \right.}{0x-4y=-12}$

$4y=-12$

Divide both sides by 4:
$\frac{4y}{4}= \frac{12}{4}$

$y=3$

Input this value into one of the Equations:
$4x+2y=18$

$4x+2(3)=18$

Simplify:
$4x+6=18$

$(4x+6)+(-6)=(18)+(-6)$

$4x=12$

Divide both sides by 4.
$\frac{4x}{4}= \frac{12}{4}$

$x=3$

The Solution to this System (The Point of Intersection):
$\boxed{(3,3)}$

I hope this helped!
-Benjamin

7 0

2 years ago

A ship sails north for 2 hours, traveling a total of 15 miles. Then the ship turns south and travels 6 miles in an hour.

ch4aika [34]

Answer:

3 miles per hour

Step-by-step explanation:

3 0

2 years ago

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

3 years ago