Match the rectangles formed by the sets of points to their corresponding areas. a(-9, 8), b(-5, 5), c(1, 13), d(-3, 16) 50 square units e(30, 20), f(39, 29), g(49, 19), h(40, 10) 300 square units i(-6, 2), j(2, 2), k(2, -8), l(-6, -8) 100 square units m(5, 5), n(11, 5), o(11, -5), p(5, -5) 80 square units q(10, 0), r(15, 5), s(25, -5), t(20, -10) u(0, 5), v(15, 20), w(25, 10), x(10, -5) arrowboth arrowboth arrowboth arrowboth

The area of ABCD is 50 units²

The area IJKL is 80 units²

The area of QRST is 100 units²

The area of UVWX is 300 units²

Step-by-step explanation:

* Lets revise the area of the rectangle

- The area of any rectangle = its length × its width

- To solve the problem find the lengths of two adjacent sides and

  consider that one of them is the length and the other is the width

- Use the rule of the distance between two points (x1 , y1) and (x2 , y2)

 the distance = √[(x2 - x1)² + (y2 - y1)²]

# In rectangle ABCD

∵ A = (-9 , 8) , B = (-5 , 5) , C = (1 , 13)

∴ AB = √[(-5 - -9)² + (5 - 8)²] = √[(4)² + (-3)²] = √[16 + 9] = √25 = 5 units

∴ BC = √[(1 - -5)² + (13 - 5)²] = √[(6)² + (8)²] = √[36 + 64] = √100 = 10 units

∴ The area of ABCD = 5 × 10 = 50 units²

# In rectangle EFGH

∵ E = (30 , 20) , F = (39 , 29) , G (49 , 19)

∴ EF = √[(39 - 30)² + (29 - 20)²] = √[9² + 9²] = √[81 + 81] = √162 unit

∴ FG = √[(49 - 39)² + (19 - 29)²] = √10² + (-10)²] = √[100 + 100] = √200 units

∴ The area of EFGH = √162 × √200 = 180 units²

# In rectangle IJKL

∵ I = (-6 , 2) , J = (2 , 2) , K = (2 , -8)

∴ IJ = √[(2 - -6)² + (2 - 2)²] = √[8² + 0²] = √8² = 8 units

∴ JK = √[(2 - 2)² + (-8 - 2)²] = √[0² + (-10)²] = √10² = 10 units

∴ The area IJKL = 8 × 10 = 80 units²

# In rectangle MNOP

∵ M = (5 , 5) , N = (11 , 5) , O = (11 , -5)

∴ MN = √[(11 - 5)² + (5 - 5)²] = √[6² + 0²] = √6² = 6 units

∴ NO = √[(11 - 11)² + (-5 - 5)²] = √[0² + (-10)²] = √10² = 10 units

∴ The area of MNOP = 6 × 10 = 60 units²

# In rectangle QRST

∵ Q = (10 , 0) , R = (15 , 5) , S = (25 , -5)

∴ QR = √[(15 - 10)² + (5 - 0)²] = √[5² + 5²] = √[25 + 25] = √50 units

∴ RS = √[(25 - 15)² + (-5 - 5)²] = √[10² + (-10)²] = √[100 + 100] = √200 units

∴ The area of QRST = √50 × √200 = 100 units²

# In rectangle UVWX

∵ U = (0 , 5) , V = (15 , 20) , W = (25 , 10)

∴ UV = √[(15 - 0)² + (20 - 5)²] = √[15² + 15²] = √[225 + 225] = √450 units

∴ VW = √[(25 - 15)² + (10 - 20)²] = √[10² + (-10)²] = √100 + 100 = √200 units

∴ The area of UVWX = √450 × √200 = 300 units²

The area of ABCD is 50 units

The area IJKL is 80 units

The area of QRST is 100 units

The area of UVWX is 300 units

Step-by-step explanation:

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: