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kirza4 [7]
3 years ago
12

Draw at least six different sized rectangles that have an area of 64 square units

Mathematics
1 answer:
olga_2 [115]3 years ago
3 0

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

A=x*y

A=64\ units^{2}

so

x*y=64 --------> equation 1

let's assume different values of x to get the different values of y

<u>case 1)</u> For x=64 units

substitute in the equation 1

x*y=64

y=64/x

y=64/64

y=1\ units      

the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

<u>case 2)</u> For x=32 units

substitute in the equation 1

x*y=64

y=64/x

y=64/32

y=2\ units

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

<u>case 3)</u> For x=16 units  

substitute in the equation 1

x*y=64

y=64/x

y=64/16

y=4\ units

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

<u>case 4)</u> For x=30 units

substitute in the equation 1

x*y=64      

y=64/x

y=64/30

y=\frac{32}{15} =2\frac{2}{15} \ units

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

<u>case 5)</u> For x=40 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/40

y=1.60\ units

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

<u>case 6)</u> For x=60 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/60

y=\frac{16}{15} =1\frac{1}{15} \ units

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

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1a.2.5x³-2.25x²+3.5x-1.5

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2a.-2x⁶+7x⁴+3x³-3x²+11x+20

2b.Yes. When multiplying, it doesn't matter which term/expression you put first

3. The equation has two solutions: x=-2, x=3

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1. (a)

(0.5x-0.25)(5x²-2x+6)

0.5x(5x²-2x+6)-0.25(5x²-2x+6)

2.5x³-x²+3x -1.25x²+0.5x-1.5

2.5x³-2.25x²+3.5x-1.5

(b)

(0.25x-0.5)(5x²-2x+6)

0.25x(5x²-2x+6)-0.5(5x²-2x+6)

1.25x³-0.5x²+1.5x -2.5x²+x-3

1.25x³-3x²+2.5x-3

This is not equal to our answer for 1(a), so we can see they're not the same.

2(a)

(-2x³+x-5)(x³-3x-4)

-2x³(x³-3x-4) +x(x³-3x-4) -5(x³-3x-4)

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-2x⁶+7x⁴+3x³-3x²+11x+20

(b)

Yes. When multiplying, it doesn't matter which term/expression you put first: xy=yx. For example, 2(3)=6 and 3(2)=6.

3

2x²-2x-12=0

(divide each term by 2)

x²-x-6=0

(x+2)(x-3)=0

Anything multiplies by zero is zero, so either x+2=0 or x-3=0

If x+2=0, x=-2. If x-3=0, x=3.

The equation has two solutions: x=-2, x=3

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