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Gnesinka [82]
2 years ago
9

How do u multiply exponents

Mathematics
2 answers:
Butoxors [25]2 years ago
8 0
Just add the powers of the exponents, and represent it by just an exponent,
ie the answer to your question remains e^11,e raise to the power of 11.
Ymorist [56]2 years ago
4 0
So, by multiplying the exponents, you would add all of the square roots together. For example for your question you would add 1+1+4+5=11 so the answer is c to the power of 11
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12 sqrt3 * 2 sqrt2 * 3

= 24 sqrt6 * 3

= 72 sqrt6  answer

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YALLL I NEED HELP !!!!!!!!!!!!!!!!!!!!!!!
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2nd one

Step-by-step explanation:

it was 0⁰F at 6 pm to -4⁰F at 11pm to you go to the left 4 times from 0⁰F and between 11pm and 5am the temperature dropped by 6 degrees so you go the the right 6 times from -4 which is 2⁰F so the 2nd one represents the temperature at 5am Hope it helped

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2 years ago
What is the side length of a square with a perimeter of 60 meters​
xxMikexx [17]

Answer:

a = 15

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Lisa sold her old bike for $140 less than she paid for it. She sold the bike for $85. Write and solve and equation to find how m
rodikova [14]

Answer:

Equation=$140+$85=$225

Step-by-step explanation:

If she sold her bike for $140 less than she paid for it, and she sold it for $85, you add 140 to 85 to get out much she paid for. To check your answer do 225-140=85.

4 0
2 years ago
Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point
Maksim231197 [3]

Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

Taking square on both the sides , we get

(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0

Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0
3 years ago
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