Which equations can be used to find the lengths of the legs of the triangle? Select three options.
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0 0.5(x)(x + 2) = 24
Ox(x + 2) = 24
O x2 + 2x – 24 = 0
O x2 + 2x - 48 = 0
0x2 + (x + 2)2 = 100

0.5(x)(x + 2) = 24 (A)

x² + 2x – 48 = 0 (D)

x² + (x + 2)² = 100 (E)

Question:

A question related to this found at (ID:4482275) is stated below.

The area of the right triangle shown is 24 square feet. Which equations can be used to find the lengths of the legs of the triangle? Check all that apply.

0.5(x)(x + 2) = 24

x(x + 2) = 24

x2 + 2x – 24 = 0

x2 + 2x – 48 = 0

x2 + (x + 2)2 = 24

Step-by-step explanation:

Find attached the diagram.

Area of triangle = ½ × base × height

= 0.5×b×h

base= x ft

Height = (x+2) ft

Area = 24ft²

24 = 0.5(x)(x+2)

0.5(x)(x + 2) = 24 (A)

The equations that can be used to find the lengths of the legs of the triangle must be equivalent to 0.5(x)(x + 2) = 24

On expanding this: 0.5(x)(x + 2) = 24

0.5(x²+2x) = 24

b) x(x + 2) = 24

x(x + 2) is not equal to 0.5(x²+2x)

c) x² + 2x – 24 = 0

0.5(x²+2x) = 24

0.5x²+x - 24 = 0 is not equal to x²+2x- 24 = 0

d) x² + 2x – 48 = 0

0.5(x)(x + 2) = 24

½(x)(x + 2) = 24

x² + 2x = 2(24)

x² + 2x – 48 = 0

Correct option (D)

x² + (x + 2)² = 100

x² + x² + 4x + 4 = 100

2x² + 4x = 96

2(x² + 2x +48)= 0

x² + 2x +48 = 0 is equal to 0.5(x²+2x) = 24

Correct (E)

the answer is d. (1.2, -4.7) on edgn

hope this !

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !