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ikadub [295]
3 years ago
6

PLS SOLVE !!!! I WILL MARK THE BRAINLIEST

Mathematics
1 answer:
laila [671]3 years ago
5 0

Answer:

A is two; B is nine; C is one

Good Luck!

a) C has to be either one or zero (but I'm assuming the solution you're wanting is when C=1). This is because A+B + Whatever is carried can only be a number 0-19, meaning only a 1 can be carried to the third column.

b) if C is 1, then in the ones place, B+A must = 11

c) if the carried 1 + a + b = 10 + a, then a can be substituted and we find that b = 9; and a =2

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The area of a rectangular pool is 8342 m .<br> If the width of the pool is 86 m, what is its length?
Minchanka [31]

Answer:

97 meters

Step-by-step explanation:

The area of a rectangle is its length multiplied by its width. Substitute the known values into this equation and solve for the length, l:

a = lw

8342 = 86l

l = 97

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A land owner is planning to build a fenced-in, rectangular patio behind his garage, using his garage as one of the "walls." He
Vitek1552 [10]

Answer:

Maximum area = 800 square feet.

Step-by-step explanation:

In the figure attached,

Rectangle is showing width = x ft and the side towards garage is not to be fenced.

Length of the fence has been given as 80 ft.

Therefore, length of the fence = Sum of all three sides of the rectangle to be fenced

80 = x + x + y

80 = 2x + y

y = (80 - 2x)

Now area of the rectangle A = xy

Or function that represents the area of the rectangle is,

A(x) = x(80 - 2x)

A(x) = 80x - 2x²

To find the maximum area we will take the derivative of the function with respect to x and equate it to zero.

A'(x)=\frac{d}{dx}(80x-2x^{2})

             = 80 - 4x

A'(x) = 80 - 4x = 0

4x = 80

x = \frac{80}{4}

x = 20

Therefore, for x = 20 ft area of the rectangular patio will be maximum.

A(20) = 80×(20) - 2×(20)²

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Maximum area of the patio is 800 square feet.

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