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Debora [2.8K]
3 years ago
8

Write 7 more than w is 25 as an algebraic equation.

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer:

w+7=25

Step-by-step explanation:

if you need it solved its

w+7=25

   -7 -7  

w=18

Thepotemich [5.8K]3 years ago
5 0

Answer:

7+w=25

Step-by-step explanation:

when it says more than, that's addition, so 7 more than w is 7+w

25 is the answer so that's self explanatory

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Part A

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The equation would be y = 12x+200 where x is the number of hours and y is the amount saved. The 200 is what you already have, and then you add on 12x additional dollars.

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Part B

  • independent variable = number of miles you drive
  • dependent variable = amount of gas in the car

equation would be y = -0.10x+12. You start off with 12 gallons and each time you use up 0.10 gallons of gas for each mile you drive, so in total you use up 0.10x gallons

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Part C

  • independent variable = number of extra credit assignments turned in
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Equation is y = 2x+60. You add on 2x additional points to the 60 you already have

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Part D

  • independent variable = number of text messages sent
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x is the number of messages and y is the total monthly bill, so the equation would be y = 0.25x + 35

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Part E

  • independent variable = number of cookies eaten per day
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In each case, the independent variable is the variable that is allowed to change and you know this value (or are given it). For instance, you know how many miles you drive and it varies. The dependent variable relies on the value chosen for the independent variable. So as x changes, then so does y.

8 0
3 years ago
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Sunny_sXe [5.5K]
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Operations with rational expressions
Scorpion4ik [409]

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

Solution:

Given expression:

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}

To solve the given expression:

First simplify: \frac{50-2 w^{2}}{3 w^{2}+9 w-30}

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}

Cancel the common factor (w + 5).

                      $=-\frac{2(w-5)}{3(w-2)}

Now substitute this in the given expression.

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}

Multiply the fractions \frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}

Factor the denominator 3(w-2)(6 w-30) =18(w-2)(w-5)

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}                        

Cancel the common factor 2(w – 5).

                               $=-\frac{w^{2}+5 w-14}{9(w-2)}

Factor the numerator w^{2}+5 w-14=(w-2)(w+7)

                               $=-\frac{(w-2)(w+7)}{9(w-2)}

Cancel the common factor (w – 2).

                               $=-\frac{w+7}{9}

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

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