Explain how the Quotient of Powers was used to simplify this expression.

2 answers:

5 0

The correct answer on how the Quotient of Powers was used to simplify the expression is D. By finding the quotient of the bases to be one-fifth and simplifying the expression. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

Ymorist [56]2 years ago

5 0

5^4
------ = 52
25

5^4
------ = 52 oh i think really this 52 mean to be 5^2
5^2

5^(4-2) = 5^2

5^2 = 5^2

hope helped

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If an object is shot with an initial velocity of 160 ft/sec, what is the smallest initial angle will you need to obtain a horizo

MrRissso [65]

To solve the question we shall use the formula for the range given by:
Horizontal range, R=[v²sin 2θ]/g
plugging in our values we get:
500=[160²×sin 2θ]/10
5000=160²×sin 2θ
0.1953=sin 2θ
thus:
arcsin 0.1953=2θ
11.263=2θ
hence:
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3 0

3 years ago

(05.05)The width of a rectangle is shown below:

Ket [755]

Answer:

no idea sorry!!

Step-by-step explanation:

i think its the third but not sure

4 0

2 years ago

A jar contains 1,200 M&M's and everyday your family eats 1/5th of the candies. After how many days will there be 314 M&M

mrs_skeptik [129]

The equation is 314 = 1,200 - 240x

1,200 is the number of M&M's it starts with
The problem said the family eats 1/5th of the candy every day
1/5 of 1,200 is 240
It is a -240 because the amount of m&m's is getting fewer each day
The x stands for the number of days

Sorry but I'm not sure if I could answer your problem exactly
The answer is 3.69 days
Because there isn't an exact day that there would exactly 314 M&M's left

7 0

2 years ago

I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?

Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Reciprocals

Algebra II

Log/Ln Property: $ln(\frac{a}{b} ) = ln(a) - ln(b)$

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

Rewrite: $t = ln(\frac{2x-1}{x-1})$
Rewrite [Ln Properties]: $t = ln(2x-1) - ln(x - 1)$
Differentiate [Ln/Chain Rule/Basic Power Rule]: $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
Simplify: $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
Rewrite: $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
Combine: $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
Reciprocate: $\frac{dx}{dt} = -(2x-1)(x-1)$
Distribute: $\frac{dx}{dt} = (1-2x)(x-1)$