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4 years ago

Cound someone please help me on this, I'm stuck.

Mathematics

1 answer:

Katyanochek1 [597]4 years ago

5 0

Negative exponents work like this:

$a^{-b}=\dfrac{1}{a^b}$

So, in order to evaluate a negative exponent, you simply have to invert the base, and then raise to the positive equivalent of the exponent.

As an example, here are the first three exercises:

$8^{-3}=\dfrac{1}{8^3}=\dfrac{1}{512}$

$(-4)^{-5}=\dfrac{1}{(-4)^5}=-\dfrac{1}{1024}$

$2k^{-4}=\dfrac{2}{k^4}$

You can work out the rest applying this logic.

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Eric has rock and classical CDs in his car. The number of rock CDs is 3 more than the number of classical CDs. Let c represent t

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Rock CD= (classical CDs + 3)
R=c+3

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3 years ago

Merrill bought a $48 sweater at sell rights department store for 10% off genie bought the same sweater at berry discount store f

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Answer: Genie paid less

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3 0

3 years ago

Given that ABCD congruent to PQRS, solve for m A.0.5 B.2.5 C.3 D.5

34kurt

As the 2 quadrilaterals are congruent the sides of one quadrilateral are equal to the corresponding sides of the other quadrilateral
Hence,
AB = PQ
RQ = BC
AD = PS
DC = RS
Since terms for both AB and PQ are given we can equate the terms and find m
5m -9 = m + 11
5m -m = 11 + 9
4m = 20
m = 5
Correct answer is D , 5

3 0

4 years ago

Bet u can’t solve this

S_A_V [24]

Answer:

Step-by-step explanation:

Assuming you're solving for p:

$m=2^{-1} *2^p(2^p-1)$

Let $y=2^p$

Now we can re-write the equation with $y$ instead of $2^p$ .

$m=\frac{1}{2} y(y-1)$

$2m=y^2-y$

$y^2-y-2m=0$

Use the quadratic formula to get:

$y = \frac{1+\sqrt{1+8m} }{2}$

$y = \frac{1-\sqrt{1+8m} }{2}$

Therefore, using natural log and log rules:

$2^p = \frac{1+\sqrt{1+8m} }{2}$ , $ln(2^p)= ln(\frac{1+\sqrt{1+8m} }{2})$ , $pln(2) = ln(\frac{1+\sqrt{1+8m} }{2})$ , $p = \frac{ln(\frac{1+\sqrt{1+8m} }{2})}{ln(2)}$

$2^p = \frac{1-\sqrt{1+8m} }{2}$ , $ln(2^p)= ln(\frac{1-\sqrt{1+8m} }{2})$ , $pln(2) = ln(\frac{1-\sqrt{1+8m} }{2})$ , $p = \frac{ln(\frac{1-\sqrt{1+8m} }{2})}{ln(2)}$