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NISA [10]
3 years ago
9

Calcula el 23 % de 175

Mathematics
1 answer:
weqwewe [10]3 years ago
5 0

Answer:40.25

Step-by-step explanation:

23% of 175

23/100 x 175

(23 x 175)/100

4025/100=40.25

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It is now twenty-seven minutes past six. What time will it be in 3 hours and 25 minutes?
stealth61 [152]

Answer:

9:52

Step-by-step explanation:

First, let's rewrite "twenty-seven minutes past six", into a standard digital clock form. We could write 6:27. Now, it's easier to see that if we add 3 hours first, we would get to 9:27. And then if we add 25 minutes, we will get to 9:52.

5 0
3 years ago
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Trey runs 3 miles in 25 minutes. At the same rate, how many miles would he run in 20 minutes?
aalyn [17]

Answer:

Step-by-step explanation:

Set up a proportion.

3/25 = x / 20      Notice that both times are in the denominators. Be sure you always set up a proportion that way. Like units in the denominator. Like units in the numerator.

Cross multiply

3 * 20 = 25 * x                        Combine

60 = 25x                                 Divide by 25

60/25 = x

x = 2.4 miles in 20 minutes.

4 0
2 years ago
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PLEASE HELP WITH MATH ASAP!!!!! ITS DUE IN 10 MINS!!! ILL GIVE BRAINLIEST
stepan [7]

Answer:

D, A, I cannot read the rest of #3, C, C.

Step-by-step explanation:

3 0
3 years ago
Help with this question please
dimaraw [331]

The answer should be C

4 0
3 years ago
A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi
viktelen [127]

Answer:

Side length = \sqrt{\frac{A}{3} } cm ,   Height =  \frac{1}{2} \sqrt{\frac{A}{3} } cm  ,  Volume = \frac{A\sqrt{A}}{6\sqrt{3} }  cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

  1. y = \frac{A - x^{2} }{4x}

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( \frac{A - x^{2} }{4x} )

V =  \frac{Ax - x^{3} }{4}

To find max volume put V' = 0

So taking derivative equation becomes

\frac{A - 3 x^{2} }{4} = 0

A = 3 x^{2}

x^{2} = \frac{A}{3}

x = \sqrt{\frac{A}{3\\} }

put value of x in equation 1

y = \frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }  

y = \frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }

y = \frac{1}{2} \sqrt{\frac{A}{3} }

So the volume will be

V = x^{2} × y

Put values of x and y from equation 2 & 3

V = \frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )

V = \frac{A\sqrt{A}}{6\sqrt{3} }

8 0
3 years ago
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