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Levart [38]
3 years ago
13

How many times 14 go into 71

Mathematics
2 answers:
suter [353]3 years ago
6 0

Answer:

5 times

Step-by-step explanation: 71 divided by 14 equals 5

coldgirl [10]3 years ago
4 0

Answer:

5 times

Step-by-step explanation:

71/14 =5r1

140=10*14

70=140/2

10/2=5

14*5=70

71-70=1

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Divide the following fractions and reduce to lowest terms 2/3 divided by 4/5
pshichka [43]
Well to divide 2 fraction you first have to find a least common number between the 2 bottom numbers in the fractions, we know automaticall that 15 fits in both the 3 and the 5. now u have to find out what to multiply by each number to get 15, for example what number do you multiply by 3 to get 15?? well it will be 5, so go ahead and multiply it by the fraction (note that what you do  in the bottom you have to do in the top) this will give you 10/15, because I multiplied the top and bottom number by 5... lets try the other one what number do you multiply by 5 to get 15?? well it will be 3 so go ahead and multiply that to get 12/15.. now add the top numbers (note that the bottoms always stay the same, you never add or subtract them) when you do you will get 22/15, now you have to see if you can simplify it any further in this case you can't so your final answer is 22/15

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3 years ago
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3 years ago
From a thin piece of cardboard 50 in. by 50 in., square corners are cut out so that the sides can be folded up to make a box. Wh
mixer [17]

Answer:

When dimension of box is 33.33 inches × 33.33 inches ×8.33  then its volume is maximum and is 9259.26 cubic inches.

Step-by-step explanation:

Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.

Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.

Thus, The dimension of box is  (50 – 2h) × (50 – 2h) × h in dimensions.

The volume V of rectangular box = (Length × Breadth × Height) cubic inches.

V=(50-2h) \times (50-2h) \times h

V=(50-2h)^2 \times h  ..............(1)

Using (a-b)^2=a^2+b^2-2ab

V=h(2500+4h^2-200h)

V=2500h+4h^3-200h^2

For obtaining a box of maximum volume, maximize V as a function of h.


Differentiate both sides with respect to h,

\frac{dV}{dh}=2500+12h^2-400h

\frac{dV}{dh}=4(625+3h^2-100h)

Solving quadratic equation,625+3h^2-100h

\frac{dV}{dh}=4(3h^2-25h-75h+625)

\frac{dV}{dh}=4(h(3h-25)-25(3h-25))

\frac{dV}{dh}=4((h-25)(3h-25))

For maximum, \frac{dV}{dh}=0  

thus,4((h-25)(3h-25))=0

⇒ h= 25 or h=\frac{25}{3}

Now check (1) for h= 25 and h=\frac{25}{3}.

h= 25 is not possible as when h is 25 inches then length and breadth becomes 0.

When h=\frac{25}{3}.

(1) ⇒ V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593  

This is the maximum volume the box can assume.

Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3  then its volume is maximum and is 9259.26 cubic inches.

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Which values are solutions to the inequality below?
RideAnS [48]
I have an answer!

ANSWER: [F] 36     Only 1 Answer!!!

No Explanation Provided!!!
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