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3 years ago

The length of a rectangle is shown below:

Mathematics

2 answers:

Natalija [7]3 years ago

7 0

Answer:

The answer is option A. C(2, −2), D(−1, −2)

Step-by-step explanation:

The length of AB = 3 units

The area of the rectangle to be drawn is 12 square units

So, the width = The area/The length = 12/3 = 4 units

C and D be located vertically 4 units below A and B

So, C and D like the image of A and B with a transformation rule

(x,y) → (x , y-4)

The coordinates of A(-1,2) and B(2,2)

So, C and D are: (-1,-2) and (2,-2)

The answer is option A. C(2, −2), D(−1, −2)

Advocard [28]3 years ago

4 0

The answer is option A. C(2, −2), D(−1, −2)

Step-by-step explanation:

The length of AB = 3 units

The area of the rectangle to be drawn is 12 square units

So, the width = The area/The length = 12/3 = 4 units

C and D be located vertically 4 units below A and B

So, C and D like the image of A and B with a transformation rule

(x,y) → (x , y-4)

The coordinates of A(-1,2) and B(2,2)

So, C and D are: (-1,-2) and (2,-2)

The answer is option A. C(2, −2), D(−1, −2)

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What number is 150% of 40?

Sedbober [7]

Answer:

150% of 40 is 60.

Step-by-step explanation:

= 150/100 × 40
= 1.5 × 40
= 60

i hope it helped. ...

8 0

3 years ago

A small tree is 80 cm tall.

ruslelena [56]

1.2 meters = 120 centimeters
amount of increase = 120-80 = 40 centimeters
40 ÷ 80 = 0.5 = 50%
There was a 50% increase.

7 0

3 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!