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LUCKY_DIMON [66]
3 years ago
12

Divide. (–6m 7 + 20m 6) ÷ 2m 4 A. –3m 3 + 20m 6 B. –3m 3 + 10m 2 C. –6m 7 + 20m 2 D. –3m 7 + 10m 6 Reset Selection

Mathematics
2 answers:
allsm [11]3 years ago
6 0
 ( –6m^7 + 20m^6<span>) ÷ 2m^4 = ???

as you know

</span> –6m^7 + 20m^6 = 2m^4 (-3m^3 + 10m^2)

[2m^4 (-3m^3 + 10m^2) ] / 2m^4 (2m^4 is canceled out)

= -3m^3 + 10m^2

answer is B. –3m^3 + 10m^2
pochemuha3 years ago
4 0

Answer:

B. -3m^7 + 10m^2

Step-by-step explanation:

First, we have to write down the division remembering that ÷ is the same as <em>/</em>

So we will write the division in terms of /

(-6m^7 + 20m^6) / 2m^4

This is the same equation as if we were using the symbol ÷

Now we have to remember the distribution when we are dividing fractions:

(a+b)/c = a/c + b/c ;

we can separate the fraction to make it easier, in this case:

<em>a = -6m^7</em>

<em>b = 20m^6</em>

<em>c = 2m^4</em>

And substituting in the fraction we have:

(-6m^7 + 20m^6) / 2m^4 = (-6m^7 / 2m^4) + (20m^6 / 2m^4)

We are going to use the second part:

(-6m^7 / 2m^4) + (20m^6 / 2m^4)

Now we are going to solve each parenthesis:

(-6m^7 / 2m^4)

To solve division that has variables with an exponent we have to remember the following:

ax^n / bx^m = (a/b) x^(n-m)

Where:

<em>a and b are constant</em>

<em>x is the variable </em>

<em>and n and m are exponents</em>

<em> </em>

In the first parenthesis (-6m^7 / 2m^4):

(a/b) x^(n-m)

(-6/2) m^(7-4)

Now we solve and we have:

(-3) m^3 this is the first part of the division

Now we have to solve the second part of the division (20m^6 / 2m^4):

(a/b) x^(n-m)

(20/2) m^(6-4)

Now we solve:

(10) m^2 this is the second part of the division

Now we have to put the two parts together:

(-3m^3) + (10 m^2)

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We can represent the Rambutan as x-4+2x or 3x-4 because it costs 2x dollars more than the Dragonfruit. 

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Therefore the overall equation would be:
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Any help with an explanation would be appreciated!
Lilit [14]

Problem 1

We'll use the product rule to say

h(x) = f(x)*g(x)

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

Then plug in x = 2 and use the table to fill in the rest

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

h ' (2) = f ' (2)*g(2) + f(2)*g ' (2)

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h ' (2) = 6 + 8

h ' (2) = 14

<h3>Answer: 14</h3>

============================================================

Problem 2

Now we'll use the quotient rule

h(x) = \frac{f(x)}{g(x)}\\\\h'(x) = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\\\\h'(2) = \frac{f'(2)*g(2)-f(2)*g'(2)}{(g(2))^2}\\\\h'(2) = \frac{2*3-2*4}{(3)^2}\\\\h'(2) = \frac{6-8}{9}\\\\h'(2) = -\frac{2}{9}\\\\

<h3>Answer:  -2/9</h3>

============================================================

Problem 3

Use the chain rule

h(x) = f(g(x))\\\\h'(x) = f'(g(x))*g'(x)\\\\h'(2) = f'(g(2))*g'(2)\\\\h'(2) = f'(3)*g'(2)\\\\h'(2) = 3*4\\\\h'(2) = 12\\\\

<h3>Answer:  12</h3>
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Answer:

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4 0
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Answer:

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