Given :
Initial length of electric cable needed, 
.
Later, Sally is told that the homeowner has decide to cut back from a three-car garage to a two-car garage, which will eliminate two branch circuits, one of 23 feet and the other of 34 feet.
To Find :
How many feet of wire will be needed now.
Solution :
Wire needed is given by :
Required = Total wire - ( length of 1nd branch + length of 2nd branch )
Required = 550 - ( 23 + 34 )
Required = 550 - ( 57 ) ft
Required = 443 ft
Therefore, length of wire required is 443 ft.
Hence, this is the required solution.
Answer:
x=15
Step-by-step explanation:
<GCB=180-5x
<EBA=9x
<DAC=8x
The sum of the three exterior angles equals 360.
(180-5x)+9x+8x=360
180-5x+9x+8x=360
-5x+9x+8x=360-180
12x=180
x=15
√45 = √9√5 = 3√5
√20 = √4√5 = 2√5
3√5+2√5-√5=4√5
answer = 4√5
:)
Answer:
Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).
So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2
f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)
f(0)=−6f(0)=−6
a(1)(−2)=−6a(1)(−2)=−6
a=3a=3
f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)
f(x)=3x3−9x−6
-12=4x-y
(-8=6x-y)-1
-12=4x-y
8=-6x+y
-4=-2x
/-2 /-2
2=x
Y=6(2)+8
Y=20
(2,20)
