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weeeeeb [17]
3 years ago
6

A club has a total of $1,520.50 in its bank account. At a budget meeting they decide to set aside 0.28 times the total amount fo

r travel. How much money will they set aside for travel? (29 points!!)
Mathematics
2 answers:
lara31 [8.8K]3 years ago
7 0

Answer:

$425.74

Step-by-step explanation:

A club has a total balance in its bank account = $1,520.50

At a budget meeting they decide to set aside 0.28 times the total amount for travel.

We have to multiply the balance to 0.28 to get the amount.

Money they will set aside for travel = 1,520.50 × 0.28

                                                          = $425.74

$425.74 they would set aside for travel.

Scilla [17]3 years ago
3 0

To solve this problem, we know from the statement that Initially, the club has $ 152050 in its bank account.

To know the amount of money you want to reserve for the trip, multiply the initial amount by 0.28.

So, we have:

$ 1,520.50 * 0.28 = $ 425.74

Finally, the club will reserve $ 425.74 from its bank account to travel.

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Step-by-step explanation:

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Solve<br><br> 4(3x - 2) = 52
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Answer:

x= 5

Step-by-step explanation:

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12x-8=52

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Vikentia [17]
So what is the question
5 0
3 years ago
1. Simplify:<br>4(4y-7y^{2})-9(5y+2)<br><br>2. Simplify:<br>24 – 4(5y – 6z) + 3y – 7z
Greeley [361]

Answer:

1)-

How to solve your question

Your question is

4(4−72)−9(5+2)

4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)

Simplify

1

Rearrange terms

4(4−72)−9(5+2)

4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)

4(−72+4)−9(5+2)

4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)

2

Distribute

4(−72+4)−9(5+2)

{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)

−282+16−9(5+2)

{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)

3

Distribute

−282+16−9(5+2)

-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)

−282+16−45−18

-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18

4

Combine like terms

2)

−17y+17z+24

See steps

Step by Step Solution:



STEP1:Equation at the end of step 1

((24 - 4 • (5y - 6z)) + 3y) - 7z

STEP2:

Final result :

-17y + 17z + 24

−282+16−45−18

-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18

−282−29−18

-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18

Solution

−282−29−18

4 0
3 years ago
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