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Flauer [41]
3 years ago
15

Jonah needs some items to repair kitchen cabinets in his home. At the Handy Man ‘R Us store he found a small can of stain for $5

.99, a tarp for $3.98, and a paint brush for $2.57. He also needs about 8 door hinges. If he only has $50.00 to spend, what is the most he can spend on each hinge? Write an inequality and solve it to answer the question.
Mathematics
2 answers:
9966 [12]3 years ago
8 0

Answer:

idk sorry

Step-by-step explanation:

Anna007 [38]3 years ago
6 0

Answer: door hinges<$4.69

Step-by-step explanation: From the question, we can deduce that

1. Jonah has just $50 and can’t spend beyond it, so his total budget must be less than or equal to $50

2. Jonah will spend a total of $12.54 (add all the items with known prices),

leaving him with $37.46(50-12.54) to spend on 8 door hinges.

Therefore 8x<=37.46(8x has to be less than or equal to $37.46)

x=door hinges

Hence, x<=4.6825 (divide both sides by 8)

But we wouldn’t spend such fraction, lol, so we’ll approximate to the nearest cent.

x<4.69 ( I chose the higher number and removed the equal sign)

OR x<=4.68

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3 years ago
Does the equation 7c+8p=29.60 help us to solve the original system? 5c+4p=18.40 2c+4p=11.20 If you think so, explain how it help
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Step-by-step explanation:

Given the simultaneous equation

5c+4p=18.40 ..... 2

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According to the equation given  7c+8p=29.60, we will see that the two equation was added to get the required equation in question. Note that this equation  7c+8p=29.60 cannot help us to solve the simultaneous equation because the resulting equation is one equation with two unknown variables. For us to get a solution to the simultaneous equation, we need to reduce the system of equation to just an equation and a variable. This can be gotten by taking the difference of the two equation as shown: (Elimination method)

Equation 1 - equation 2 will give;

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p = -4.27/2

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You can see that what helped in solving the system of equation is by eliminating one of the variables first using the Elimination method.

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