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Nady [450]
4 years ago
12

Which expressions are equivalent to the one below? Check all that apply.

Mathematics
2 answers:
d1i1m1o1n [39]4 years ago
6 0

Use two main <u>properties</u> for <u>powers</u>:

1. (a^m)^n=a^{m\cdot n};

2. \dfrac{a^m}{a^n}=a^{m-n}.

The expression \dfrac{25^x}{5^x} can be simplified in following way:

\dfrac{25^x}{5^x}=\dfrac{(5^2)^x}{5^x}=\dfrac{5^{2x}}{5^x}=5^{2x-x}=5^x.

Then the expression \dfrac{25^x}{5^x} is equivalent to 5^x.

Answer: correct choice is D

In-s [12.5K]4 years ago
3 0
The answer is E:5
a^n/b^m=a/b^n/m
So x/x=1 and 25/5=5
Then 5^1=5
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How many 100 make 3200?
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32!

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Write an equation of the line passing through the point A(2, 0) that is parallel to the line y = 3x−5 .
sammy [17]
Hello,

In order for lines to be parallel to each other, they need to have the same slope. So, let's plug in the points and slope for point-slope form.

The equation for point-slope form is: y-y1 = m (x-x1)

Let's plug in the points into the equation. 
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After plugging the numbers in, the equation is simplified to y-5 = 3 (x-0). Then distribute the 3 to the parenthesis. y-5 = 3x. To get y by itself, you need to add 5 to both sides. After that, the equation parallel to that equation is: y = 3x + 5.

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Using the letters in the word INNOVATIVE, find the number of permutations that can be formed using 4 letters at a time. Show you
lyudmila [28]

Answer:  1) 5040 and 2) 165

Step-by-step explanation:

1) Here total number of letters = 10

The number of permutations that can be formed using 4 letters at a time

= P (10, 4)

= 10_P_4

= \frac{10!}{(10-4)!}

=  \frac{10!}{6!}

= \frac{10\times 9\times 8\times 7\times 6!}{6!}

= 10 × 9 × 8 × 7

= 5040

2) Here the total number of machine = 11

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= \frac{11!}{(11-3)!3!}

= \frac{11!}{8!3!}

= \frac{11\times 10\times 9\times 8!}{8!\times 6}

= 11 × 5 × 3

= 165

5 0
3 years ago
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