Subtract 7x on both sides of the equal sign (it'll cancel out) then -3=-3 which makes a true statement.
Answer:
14
Step-by-step explanation: 2(4x+6)=
\,\,34-3x
34−3x
Since HJ is a midsegment, twice HJ is equal to EG.
8x+12=
8x+12=
\,\,34-3x
34−3x
Distribute.
8x+12=
8x+12=
\,\,-3x+34
−3x+34
Communicative property to change the order
\color{red}{+3x}\phantom{+12}\phantom{=}
+3x+12=
\,\,\color{red}{+3x}\phantom{+34}
+3x+34
+3x to both sides.
11x+12=
11x+12=
\,\,34
34
\phantom{11x}\color{red}{-12}\phantom{=}
11x−12=
\,\,\color{red}{-12}
−12
-12 to both sides.
11x=
11x=
\,\,22
22
\frac{11x}{11}=
11
11x
=
\,\,\frac{22}{11}
11
22
Divide both sides by 11
x=
x=
\,\,2
2
Value of x
HJ=
HJ=
\,\,4x+6
4x+6
Value of HJ
HJ=
HJ=
\,\,4(2)+6
4(2)+6
Plug in x.
HJ=
HJ=
\,\,8+6
8+6
Multiply.
HJ=
HJ=
\,\,14
14
Answer:
a) 13 m/s
b) (15 + h) m/s
c) 15 m/s
Step-by-step explanation:
if the location is
y=x²+3*x
then the average velocity from 3 to 7 is
Δy/Δx=[y(7)-y(3)]/(7-3)=[7²+3*7- (3²+3*3)]/4= 13 m/s
then the average velocity from x=6 to to x=6+h
Δy/Δx=[y(6+h)-y(6)]/(6+h-6)=[(6+h)²+3*(6+h)- (6²+3*6)]/h= (2*6*h+3*h+h²)/h=2*6+3= (15 + h) m/s
the instantaneous velocity can be found taking the limit of Δy/Δx when h→0. Then
when h→0 , limit Δy/Δx= (15 + h) m/s = 15 m/s
then v= 15 m/s
also can be found taking the derivative of y in x=6
v=dy/dx=2*x+3
for x=6
v=dy/dx=2*6+3 = 12+3=15 m/s
Answer:
(c, d) = (25, 35)
Step-by-step explanation:
Multiply the first equation by 2.5 and subtract the second one:
2.5(c +d) -(2.5c +1.75d) = 2.5(60) -(123.75)
0.75d = 26.25 . . . . . . . . . simplify
26.25/0.75 = d = 35 . . . . divide by the coefficient of d
60 -d = c = 25 . . . . . . . . . use the first equation to find c
(c, d) = (25, 35)
