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otez555 [7]
3 years ago
12

Find the greatest common factor of 12m^2 and 9b^4

Mathematics
1 answer:
mylen [45]3 years ago
8 0

Answer:

The greatest common factor of 12m^2 and 9b^4 is 3.

I hope this could help, let me know if you need anymore help.

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Pease answer quick (only 9.1)
Dafna11 [192]

Answer:

32

24

18

13.5

..............................

6 0
3 years ago
Jennifer Drive 265 miles in five hours what is her average rate of speed and miles per hour
iris [78.8K]

Answer:

53mph

Step-by-step explanation:

Divide 265 by 5, there is your answer.

3 0
3 years ago
Read 2 more answers
Jade wants to buy a $200,000 term life insurance policy. She is 34 years old. Using the premium table, what is her annual premiu
AnnZ [28]

Premium rates are usually for  premium per fixed face value. Jade's annual premium for a 10 year policy is given by Option b: $1,202

<h3>How to calculate the total annual premium for $x ?</h3>

If its given that the annual premium is $p per $y face value, then we can calculate the annual premium for $1 face value and then use it to calculate annual premium for $x.

Using proportions, we get:

\rm \$y \: face \: value : \$p \: annual \: premiun\\\\\rm \$1 \: face \: value : \$\dfrac{p}{y} \: annual \: premiun\\\\\rm \$x\:face\: value : \$\dfrac{p \times x}{y} \: annual \: premiun\\

For given case, from the tables, we see that for age 34, and 10 year life insurance for female gender , there is annual premium of 6.01 per $1000 face value.
Thus, we have p = 6.01, y = 1000

Since Jade wants to buy Life insurance for $200,000, thus, x = $200,000

Putting it in the above derived formula, we get:

\rm \$x\:face\: value : \$\dfrac{p \times x}{y} \: annual \: premium\\\\\rm \$200000\:face\: value : \$\dfrac{6.01 \times 200000}{1000} \: annual \: premium\\ = \$1202 \: annual \: premiun

Thus, Jade's annual premium for a 10 year policy is given by Option b: $1,202

Learn more about calculating annual premium cost here:

brainly.com/question/13168988

4 0
3 years ago
Given the coordinates below, what is the midpoint of AB?<br> A(-6, -10)<br> B(2,5)
madreJ [45]

Given :

\:

  • A(-6, -10)
  • B( 2, 5)

\:

\gray{ \frak{The \:  given \:  two \:   \: points  \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}

\:

Let's solve by using midpoint formula :

\:

\bf \boxed{\color{red}\frak{Midpoint \:   \: Formula :  {( \: x ,\:y \:  ) = }(  \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}

\:

\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}

\:

\: \large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}

\:

\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{  \cancel\frac{ -4}{2} ,  \cancel\frac{ - 5}{2} )}}

\:  \:

\underline{ \boxed{ \large \red{ \frak{option \: d }(  \frak{   - 2, - 2.5)}}}}✓

Hope Helps! :)

6 0
2 years ago
Simplify square root of 1008 ( V` Represents the square root sign.)
Alexus [3.1K]
So first factor it
1008=2*2*2*2*3*3*7
group them in similar pairs
1008=(2*2)(2*2)(3*3)*7
now remember we can do
\sqrt{xy}= \sqrt{x} \sqrt{y} so
\sqrt{1008}= (\sqrt{2*2}) (\sqrt{2*2}) (\sqrt{3*3}) (\sqrt{7})=
2*2*3*\sqrt{7}=
12√7

the answer is B

5 0
3 years ago
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