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Ksju [112]
2 years ago
5

What is 6x + 4x? Explain your reasoning.

Mathematics
1 answer:
Tamiku [17]2 years ago
5 0

Answer:

10x

Step-by-step explanation:

Because if you add 6 and 4 it becomes 10 and there is x with 6 and 4 both so it is 10x. Whereas x= - 10

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I have no clue what to do
igor_vitrenko [27]

Answer:

4x + 3y = 12

Step-by-step explanation:

6 0
1 year ago
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Can someone do this for me?
Amanda [17]
Are theese questions or your  answers

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7 0
3 years ago
Barney & Noblet customers can choose to purchase a membership for $25 per year Members receive 10% off all store purchases
Law Incorporation [45]

Answer:

  1. $70
  2. y = 25 + 0.9x
  3. $250

Step-by-step explanation:

1. 10% of $50 is $5, so the purchases would come to $50 -5 = $45. Added to the $25 membership fee, the total cost for the year would be

  $45 +25 = $70

2. The member pays $25 even if no purchases are made. Then any purchases are 100% - 10% = 90% of the marked price. So, the total is ...

  y = 25 + 0.90x

3. $25 is 10% of $250, so that is the amount the member would have to purchase to break even on cost.

If you like, you can compare the cost without the membership (x) to the cost with the membership (25+.9x) and see where those costs are equal.

  x = 25 +0.9x . . . . . x is the spending level at which there is no advantage

  0.1x = 25 . . . . . . . . subtract 0.9x

  25/0.1 = x = 250 . . . divide by 0.1

4 0
3 years ago
Given the data presented in the bar graph, which age group represents 25% of the people at the family reunion? A) 10−19 B) 20−29
Dmitriy789 [7]

Answer:

B

Step-by-step explanation:

First you find the amount of people at the reunion which was 80. Then you find 25% of 80. 25% of 80 is 20. The only age group with 20 people is the 20-29 year olds.

4 0
2 years ago
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Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
Whitepunk [10]

If f(x)=4x-3:

\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4

If f(x)=4x^{-3}:

\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}

\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}

\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}

7 0
3 years ago
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