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Zanzabum
3 years ago
10

Simplify. Write the expression using only positive exponents. m^−2 times n^3

Mathematics
2 answers:
steposvetlana [31]3 years ago
8 0
N^3/m^2 I believe is what you’re looking for
Sladkaya [172]3 years ago
8 0

Answer:

\boxed{\bold{\frac{n^3}{m^2}}}

Step By Step Explanation:

  • Apply Exponent Rule \bold{\:a^{-b}=\frac{1}{a^b}}

\bold{m^{-2}=\frac{1}{m^2}}

\bold{n^3\frac{1}{m^2}}

  • Multiply Fractions \bold{\:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}}

\bold{\frac{1\cdot \:n^3}{m^2}}

  • Multiply: \bold{1\cdot \:n^3=n^3}

\bold{\frac{n^3}{m^2}}

\boxed{\bold{Mordancy}}

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If this is true then what does K equal to? Write as an inequality. plz help me. you will be greatly appreciated.
Scrat [10]

Answer:

K = (2v)/2

Step-by-step explanation:

1. Multiply both sides by 2/n

<u>Now: (2v)/n = b + K</u>

<u></u>

2. Subtract k from both sides

<u>Now: (2v)/n - b = K</u>

8 0
3 years ago
Find the 5th term of the sequence defined by the give rule. f(n) = n²+ 5. A step by step explanation with answer would be greatl
skad [1K]

Answer:

<h3>The 5th term is 30</h3>

Step-by-step explanation:

Given the formula

f(n) = n² + 5

where n is the number of terms

So from the question we were told to find the 5th term that's

n = 5

In order to find the 5th term substitute the value of n that's 5 into the rule

We have

f(5) = 5² + 5

= 25 + 5

= 30

<h3>f(5) = 30</h3>

So the 5th term of the sequence in the given rule is 30

Hope this helps you

3 0
4 years ago
PLEASE RESPOND FAST!! a number cube has sides numbered 1 through 6 the probabilty of roling a 2 is what is the probabilty of not
zubka84 [21]

Answer: 5/6

Step-by-step explanation:

5 0
3 years ago
What is the value of side b?<br> A 20<br> B 15<br> C 10<br> D 5
Pavel [41]
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3 0
3 years ago
Read 2 more answers
The output of an economic system​ Q, subject to two​ inputs, such as labor L and capital​ K, is often modeled by the​ Cobb-Dougl
Anna35 [415]

Answer:

Step-by-step explanation:

Given

Economic system Q is given by

Q=cL^aK^b

also a+b=1

if Q=12,200

a=\frac{1}{6}

b=\frac{5}{6}

c=42

substitute these values

12,200=42\times (L)^{\frac{1}{6}}K^{\frac{5}{6}}

(L)^{\frac{1}{6}}K^{\frac{5}{6}}=\frac{12,200}{42}

K^{\frac{5}{6}}=\frac{12,200}{42(L)^{\frac{1}{6}}}

K=(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^{5}}

differentiate w.r.t to L to get \frac{dK}{dL}

\frac{dK}{dL}=(\frac{12,200}{42})^{\frac{6}{5}}\times (-5)\times L^{-6}

\frac{dK}{dL}=-5(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^6}

\frac{dK}{dL}=-\frac{4515.466}{L^6}

6 0
3 years ago
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