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Kipish [7]
2 years ago
10

Beth made wristbands and belts for a craft sale. She sold 30 of these items. Each Wristband sold for $5.50. Each belt sold for $

8.75. It’s Beth made $204 at the craft sale, how many wristbands did she sell? How many belts did she sell? Write and solve a system of equations to solve the problem. Show your work.
Mathematics
1 answer:
Serhud [2]2 years ago
3 0
Let W = number of wristbands.
Let B = number of belts.

Number equation: W + B = 30
Value equation: 5.50W + 8.75B = 204.00

Multiply the top equation by 5.50 and subtract it from the bottom equation:
5.50W + 8.75B = 204.00 <--- 5.50(W + B = 30)
5.50W + 5.50B = 165.00
---------------------------------
---> 3.25B = 39.00 <--- subtract down the columns
---> B = 12 <--- divide by 3.25

Since W + B = 30 ---> W + 12 = 30 ---? W = 18

There were 12 belts and 18 wristbands.
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<em>The difference of, the product of three and n squared, and x.</em>

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3 years ago
Mariam’s uncle donate 120 cans of juice and 90 packs of cheese cracker for the school picnic. Each student must receive the same
monitta

Answer:

8 students,

15 cans of juice and 11 packs of cheese cracker

Step-by-step explanation:

Mariam’s uncle donate 120 cans of juice and 90 - 2 = 88 packs of cheese cracker (eats two packs) for the school picnic.

Factor these numbers:

120=2\cdot 60=2\cdot 2\cdot 30=2\cdot 2\cdot 2\cdot15=2\cdot 2\cdot 2\cdot 3\cdot 5\\ \\88=2\cdot 44=2\cdot 2\cdot 22=2\cdot 2\cdot 2\cdot 11

The greatest common factor is

2\cdot 2\cdot 2=8,

so the greatest number of students who can share the food equally is 8 and each student will get 3\cdot 5=15 cans of juice and 11 packs of cheese cracker.

4 0
3 years ago
Plsss hurry i need help brainliest involed
dybincka [34]
45.5 is your answer, you will have to multiple the height times the base ( 5 x 9.1 )
7 0
2 years ago
You need to solve a system of equations. You decide to use the elimination method. Which of these is not allowed?
Serggg [28]

Answer:

B

Step-by-step explanation:

The <u>Elimination Method</u> is the method for solving a pair of linear equations which reduces one equation to one that has only a single variable.

  • If the coefficients of one variable are opposites, you add the equations to eliminate a variable, and then solve.
  • If the coefficients are not opposites, then we multiply one or both equations by a number to create opposite coefficients, and then add the equations to eliminate a variable and solve.

When multoplying the equation by a coefficient, we multiply both sides of the equation (multiplying both sides of the equation by some nonzero number does not change the solution).

So, option B is not allowed (it is not allowed to multiply only one part of equation)

8 0
3 years ago
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