Answer the 3 question with the Task card Thank you.

Please see below.

Step-by-step explanation:

I will be using * to express multiplication, and ^ to express squares.

A) Can you find the area of the shape?

Yes, we can.

We can simply multiply 5x - 3 by 2x, as seen below.

(2x)(5x - 3)

2x * 5x = 10x^2

2x * -3 = -6x

Area = 10x^2 - 6x

B) Can you find the perimeter of the shape?

Yes.

Because a rectangle has two sets of equal sides, we can merely multiply the two given measurements by two and add them together to find the perimeter.

2x * 2 = 4x

2(5x - 3) = 10x - 6

4x + 10x - 6 = 14x - 6

Perimeter = 14x - 6

C) If x = 2, what is the area of the shape?

We can just plug in 2 wherever our x shows up.

5x - 3 = 5 * 2 - 3 = 10 - 3 = 7

2x = 2 * 2 = 4

7 x 4 = 28

We can also use our previous formula from A.

10x^2 - 6x

10 * 2^2 - 6 * 2

10 * 4 - 12

40 - 12

28

Our area if 28 if x = 2

Please let me know if any of these are incorrect.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18