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3 years ago

Larry bought 24 marbles little lost half of his marbles how many did he lose

Mathematics

1 answer:

emmasim [6.3K]3 years ago

3 0

Answer:

Step-by-step explanation:

his he had lost 12 of his marbles then he would only have 12 left to begin with.

to do this we can do the equation:

24/0.5 (the 0.5 is the represent that the 24 marbles had been split into two, a half for each party)

this would then give use our quotient of 12 marble he had lost

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Cindy has a chocolate box whose length is 12 cm, height of 8 cm, and width of 6 cm. Find the volume of the box?

DerKrebs [107]

Hello there I hope you are having a great day :) Cindy has a chocolate box whose length is 12 cm, height of 8 cm, and width of 6 cm. Find the volume of the box?

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2 years ago

Read 2 more answers

You can buy 5 cans for green beans at the Village Markets for $3.70. You can buy 10 of the same cans of beans at Sam’s Club for

mihalych1998 [28]

The place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.

Village market:

Green beans = 5 cans
Total cost = $3.70

Unit rate = Total cost / green beans

= 3.70 / 5

= $0.74 per can

Sam's club:

Green beans = 10 cans
Total cost = $5.90

Unit rate = Total cost / green beans

= 5.90/10

= $0.59 per can

Therefore, the place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.

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brainly.com/question/4895463

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1 year ago

Maria has $28 to spend on bottles of water Each bottle of water costs $1.29. There is a $1.50 recycling fee for buying bottled w

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4 0

3 years ago

Worth 15 points please answer asap!!!!

ExtremeBDS [4]

So do the opposite of the answer like for example 4 +3\2 and your answer is y
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3 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

8 0

1 year ago

Larry bought 24 marbles little lost half of his marbles how many did he lose​

Larry bought 24 marbles little lost half of his marbles how many did he lose