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Nitella [24]
3 years ago
6

Mrs. Kellen's most recent math test consisted of 50 questions. Some of the questions were worth two points and the rest of the q

uestions were worth 3 points. If there were 115 points available on the test, how many questions were worth 2 points?
Mathematics
1 answer:
WARRIOR [948]3 years ago
7 0

Answer:

The answer to your question is: 35 questions worth 2 points and 15 questions worth 3 points.

Step-by-step explanation:

Data

Number of questions = 50

Total points = 115

x = questions worth 2 points

y = questions worth 3 points

Process

write to equations

(1)  ---------------    x + y = 50

(2) --------------    2x + 3y = 115

Solve them by substitution

           x = 50 - y

         2 (50 - y) + 3y = 115

          100 - 2y + 3y = 115

           3y -2 y = 115 - 100

                     y = 15

           x = 50 - 15

            x = 35

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m_{perpendicular} = - \frac{1}{m}

the given slope m = 4, hence

m_{perpendicular} = - \frac{1}{4}


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Step-by-step explanation:

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Answer:

The filled table for each equation by using the exact values in the table is

10x+2y=56                                          

x                                           y

______________________    

0                                          28

\frac{56}{10}                                          0

________________________

8x+3y=49

x                                               y

_________________________

0                                             \frac{49}{3}

\frac{49}{8}                                           0

_________________________

Step-by-step explanation:

Given equations are 10x+2y=56 and 8x+3y=49

To fill the table for each equation by using the exact values in the table :

10x+2y=56

put x=0 in above equation we get

10(0)+2y=56

2y=56

y=\frac{56}{2}

y=28

Therefore (0,28)

put y=0 in the given equation 10x+2y=56 we get

10x+2(0)=56

10x=56

x=\frac{56}{10}

Therefore (\frac{56}{10},0)

10x+2y=56

x                                               y

_________________________

0                                             28

\frac{56}{10}                                              0

__________________________

For

8x+3y=49

put x=0 in above equation we get

8(0)+3y=49

3y=49

y=\frac{49}{3}

Therefore (0,\frac{49}{3})

put y=0 in the given equation 8x+3y=49 we get

8x+3(0)=49

8x=49

x=\frac{49}{8}

Therefore (\frac{49}{8},0)

8x+3y=49

x                                               y

_________________________

0                                             \frac{49}{3}

\frac{49}{8}                                             0

________________________

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