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3 years ago

Hey can you please answer this question on 7th grade exponents? thank you! boo

Mathematics

2 answers:

Ket [755]3 years ago

8 0

A. 75 x 0.0016= .12 Has a value between 0 and 1

B. 3^5 / 3^-6 = 177147 Does not a value between 0 and 1

C. 1/4^3 x 1/4^2 = about 0.0009 Has a value between 0 and 1

D. -7^5 / -7^7 = about 0.02 Has a value between 0 and 1

s2008m [1.1K]3 years ago

5 0

I did this before and it’s b

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elena-s [515]

Answer:

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Step-by-step explanation:

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2 years ago

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Perry is buying an I phone that is a sale for 40% off the original price.after he uses a $25 gift card.the total cost before tax

marta [7]

Answer:

$599

Step-by-step explanation:

Given: Discount- 40% off on original price.

Gift card= $25

Total cost before taxes is $334.40

Lets assume the original price be "x".

We know there is 40% on original price and Perry uses $25 gift card.

Remember; Total cost= $Original\ price- discount- gift\ card$

Now, finding the original price of Iphone.

⇒ $x-0.40x-25= 334.40$

⇒ $0.60x-25= 334.40$

Adding both side by 25

⇒ $0.60x= 334.40+25$

⇒ $0.60x= 359.4$

Dividing both side by 0.60

⇒ $x= \frac{359.4}{0.60}$

∴ $x= 599$

Hence, the original price of iphone is $599

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3 years ago

How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t

Zarrin [17]

By first principles, the derivative is

$\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h$

Use the binomial theorem to expand the numerator:

$(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n$

$(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

The first term is eliminated, and the limit is

$\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h$

A power of $h$ in every term of the numerator cancels with $h$ in the denominator:

$\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)$

Finally, each term containing $h$ approaches 0 as $h\to0$ , and the derivative is

$y=x^n\implies y'=nx^{n-1}$

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3 years ago

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snow_tiger [21]

Answer:

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Step-by-step explanation:

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2 years ago

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Blababa [14]

Answer:

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2 years ago