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3 years ago

Which is a possible expression for the series -6-12-24-48-96-192

Mathematics

1 answer:

iren [92.7K]3 years ago

4 0

Answer:

Math is confusing

Step-by-step explanation:

LOL

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Question 2

Free_Kalibri [48]

I’m terms of 12 in is the correct answer.

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2 years ago

A friend provides 100 pieces of supporting evidence for a mathematical statement. What is the minimum number of counterexamples

seraphim [82]

Answer:

1 true good statement can prove his friend wrong.

Step-by-step explanation:

Answer is 1.

7 0

2 years ago

Help me please djsjns

Svetach [21]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

I need help!!

mestny [16]

I think the answer is +4

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3 years ago

Pleases someone explain

Nikolay [14]

When you have a negative exponent, you move the base with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

$\frac{1}{y^{-2}} =\frac{y^2}{1}$ or $y^2$

$2x^{-3}$ or $\frac{2^1x^{-3}}{1} =\frac{2}{x^3}$

$\frac{1}{x^{-5}} =x^5$

When a base with an exponent is divided by a base with an exponent, you subtract the exponents together. (But you can only combine the exponents when the bases are the same)

For example:

$\frac{x^2}{y}$ (can't combine because they have different bases of y and x)

$\frac{x^5}{x^3} =x^{(5-3)}=x^2$

$\frac{2^2}{2} =2^{(2-1)}=2^1 = 2$

When you multiply an exponent directly to a base with an exponent, you multiply the exponents together.

For example:

$(x^2)^3=x^{(2*3)}=x^6$

$(y^2)^{10}=y^{(2*10)}=y^{20}$

$\frac{2e^0}{(e^{-3})^2}$ First multiply the exponents together in the denominator

$\frac{2e^0}{e^{(-3*2)}}= \frac{2e^0}{e^{-6}}$ Now subtract the exponents together

$2e^{(0-(-6))}$ (two negative signs cancel each other out and become positive)

$2e^{(0+6)}$

$2e^6$

7 0

3 years ago