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Dvinal [7]
2 years ago
8

Ade is baking mini-loaves of bread. How many mini-loaves could he make with 10 cups of flour?

Mathematics
1 answer:
BabaBlast [244]2 years ago
5 0

Answer: 25

Step-by-step explanation:

Since Ade uses 2 cups of flours for 5 mini loaves, it simply means that 1 cup of flour will be used for:

= 5/2 = 2.5 mini loaves

Therefore, the number of mini-loaves that he could make with 10 cups of flour would be:

= 10 × 2.5

= 25 mini loaves

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Write the numbers 23,643 and 23,987 so they line up by place value. Explain how to line them up
weqwewe [10]

Consider the numbers 23,643 and 23,987.

Since both the given numbers have same digits at ten thousands and thousands place.

So, we can'not compare the numbers on the basis of ten thousands and thousands digit.

So, let us consider the number 23,643

Consider the place value of digit at the hundreds place. Since, the digit at hundreds place = 6.

The place value of 6 = 600

Now, let us consider the number 23,987

Consider the place value of digit at the hundreds place. Since, the digit at hundreds place = 9.

Te place value of 9 = 900

Since, 900 > 600

So, 23,643 > 23,987

Therefore, the number 23,987 is greater than the number 23,643.

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3 years ago
A pen is being built that is 10 feet long and 20 feet wide. How much fencing material will be needed?
mr_godi [17]
Perimeter = 2(10+20) = 60feet

60ft fencing material is needed
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Find the area of the composite figure
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Well cutting off the small 4x2 you get 8 plus’s the 7x2 is 14 so just add those together you should get 22in ^2
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Mick paid 2.94 in sales tax on an item that cost $42.00 before tax.at that rate how much would he pay in sales tax for an item t
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A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. If section C held 300 fewer
White raven [17]

Answer:

Let's define the variables:

A = number of seats in section A.

B = number of seats in section B.

C = number of seats in section C.

We have the equations:

A + B + C = 45,000.

C - 300 = B/2

A*$28 + B*$24 + C*$20 = $1,139,200

This is a system of equations, the first step to solve this is to isolate one variable in one of the equations, and then replace it in the others.

I will isolate C in the second equation:

C = B/2 + 300.

Now let's replace this in the other two equations:

A + B + B/2 + 300 = 45,000

A*$28 + B*$24 + (B/2 + 300)*$20 = $1,139,200

Let's simplify these equations:

A + B*(3/2) = 44,700

A*$28 + B*$34 + $6,000 = $1,139,200

Now let's isolate A in the first equation:

A = 44,700 - B*(3/2)

Let's replace this in the other equation:

(44,700 - B*(3/2))*$28 + B*$34 + $6,000 = $1,139,200

Now let's solve this for B.

-B*$8 + $1,252,200 = $1,139,200

-B*$8 =  $1,139,200 - $1,252,200  = -$113,000

B = $113,000/8 = 14,125

Now we can replace that in the equations:

A =  44,700 - B*(3/2) =  44,700 - 14,125*(3/2) = 23,512.5, that we should round up to 23,513.

And C = B/2 + 300 = 7362.5

As we rounded the previous one up, we should round this one down to 7362.

3 0
2 years ago
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