AK = 640
Δ ABC and ΔFJK are similar. They are small triangles.
ΔCDF is the big triangle.
640 / 2 = 320 m= CF
AC & FK= 320/2 = 160 m each
2AC = CF = 2FK
2(160) = 320 = 2(160)
BG = 20 m
20/160 = x / 320
20*320 = 160x
6400 = 160x
6400/160 = x
40 = x
Area of CDF = (40 m * 320 m)/2 = 12,800 / 2= 6,400 m²
Answer:
x = 3 or x = 7
Step-by-step explanation:
X² - 8x + 14 = 2x - 7
x² - 8x + 14 - 2x + 7 = 0
Now we have a quadratic equation ;
x² - 10x + 21 = 0
-3 and - 7 gives a sum of - 10 and a product of 21
x² - 3x - 7x + 21 = 0
x(x - 3) - 7(x - 3) = 0
(x - 3) = 0 or (x - 7) = 0
x = 3 or x = 7
The Statement would be: 0.16 < 0.62
Area of the trapezoid = 1/2(B+b)h
where
B= length of the longer side of the trapezoid which is equal to 14 ft
b= shorter shorter side of the trapezoid which equal 8 ft
h = height of the trapezoid which is equal to 4 ft
Area of the trapezoid = 1/2 (14+8)4
Area of the trapezoid yard fence of Duc is 44ft^2
Answer: <em>This quadratic equation has two real solutions.</em>
<em />
Step-by-step explanation:
