Answer:
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 brainly (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)
Answer:
x+8=-3 gives
x=-11 is the required value
Answer:do h just want answer or explanation
Step-by-step explanation:?
add 4 i guess. i dont really know
Quartic is 4th degree
the factors of an equation with roots r1,r2 is
(x-r1)(x-r2)
4th degree
it could be
(x-r1)¹(x-r2)³ or
(x-r1)²(x-r2)² or
(x-r1)³(x-r2)¹
roots or zeroes at x=-1 and x=-2
(x-(-1)) and (x-(-2))
(x+1) and (x+2)
the function could be factored into
(x+1)¹(x+2)³ or
(x+1)²(x+2)² or
(x+1)³(x+2)¹
expanded would be
x⁴+7x³+18x²+20x+9 or
x⁴+6x³+13x²+12x+4 or
x⁴+5x³+9x²+7x+2
one of those is the answer