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2 years ago

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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Irina18 [472]

12 sqrt3 * 2 sqrt2 * 3

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3 years ago

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Step-by-step explanation:

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2 years ago

What is the side length of a square with a perimeter of 60 meters

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Step-by-step explanation:

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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.

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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.

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3 years ago