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Volgvan
4 years ago
14

Line m has no y-intercept, and its x-intercept is (3, 0). Line n has no x-intercept, and its y-intercept is (0, -4).

Mathematics
2 answers:
Makovka662 [10]4 years ago
7 0

for line m it should be y=3x+0

line n should be y=0x+-4

kaheart [24]4 years ago
3 0

Answer:

m is x=3

n is y=-4

Step-by-step explanation:

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3 years ago
The Social Justice Club is going to rent a mammoth barbeque to cook the hot dogs for the hot dog sale After contacting three ren
zimovet [89]

Answer:

(A)Cost of Rental A, C= 15x

Cost of Rental B, C=5x+50

Cost of Rental C, C=9x+20

(B)

i. Rental C   ii. Rental A     iii. Rental B

Step-by-step explanation:

<u>Part 1</u>

Let x be the number of hours of the barbeque use by the club.

Rental A: $15/h

Cost of Rental A, C= 15hx

Rental B: $5/h + 50

Cost of Rental B, C=5x+50

Rental C: $9/h + 20

Cost of Rental C, C=9x+20

<u>Part 2</u>

The graph of the three models is attached below

<u>Part 3(11.05-4.30)</u>

If the barbecue's usage hour is from 11.05 to 4.30 when the football match ends.

Number of Hours between 11.05am and 4.30pm=4 hours 25 Minutes = 4.42 Hours

Cost of Rental A, C= 15x=15(4.42)=$66.30

Cost of Rental B, C=5x+50 =5(4.42)+50=$72.10

Cost of Rental C, C=9x+20=9(4.42)+20=$59.78

Rental C should be chosen as it offers the lowest cost.

<u>Part 4 (11.05-12.30)</u>

Number of Hours = 12.30 -11.05 =1 hour 25 Minutes = 1.42 Hours

  • Cost of Rental A, C= 15x=15(1.42)=$21.30 Cost of Rental B, C=5x+50 =5(4.42)+50=$57.10 Cost of Rental C, C=9x+20=9(4.42)+20=$32.78

Rental A should be chosen since it offers the lowest cost.

<u>Part 5</u>

If the barbecue is returned the next day, say after 24 hours

  • Cost of Rental A, C= 15x=15(24)=$360 Cost of Rental B, C=5x+50 =5(24)+50=$170 Cost of Rental C, C=9x+20=9(24)+20=$236

Rental B should be chosen as it offers the lowest cost.

6 0
3 years ago
1/3y+11=1/2y-3<br><br> What is y???
JulijaS [17]

Answer:

y = 84

Step-by-step explanation:

1) add 3 to both sides of your equation to cancel out the -3

end up with: (1/3)y + 14 = (1/2)y

2) multiply both sides of your equation by 2 to cancel out the (1/2)

end up with: (2/3)y + 28 = y

3) subtract (2/3)y from both sides of the equation to cancel out the positive (2/3)y

end up with: 28 = (1/3)y

4) multiply both sides of the equation by 3 to cancel out the (1/3)

end up with: 84 = y

6 0
4 years ago
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