Answer:
a. 14 is a coefficient, and b is a variable.
This can be expanded to 14*b
b. 30 is a coefficient, and j and k are variables.
This can be expanded to 30*j*k
Answer: OPTION A.
Step-by-step explanation:
You can observe that in the figure CDEF the vertices are:

And in the figure C'D'E'F' the vertices are:

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:
For C'(-8,-4) and C(-2,-1):

Let's choose another vertex. For E'(8,8) and E(2,2):

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.
Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:
→
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
Answer: The dividend is divided by the divisor to get the result which is the quotient.
A
Step-by-step explanation:First, subtract
2
π
r
2
from each side of the equation to isolate the
h
term:
S
−
2
π
r
2
=
2
π
r
h
+
2
π
r
2
−
2
π
r
2
S
−
2
π
r
2
=
2
π
r
h
+
0
S
−
2
π
r
2
=
2
π
r
h
Now, divide each side of the equation by
2
π
r
to solve for
h
:
S
−
2
π
r
2
2
π
r
=
2
π
r
h
2
π
r
S
−
2
π
r
2
2
π
r
=
2
π
r
h
2
π
r
S
−
2
π
r
2
2
π
r
=
h
h
=
S
−
2
π
r
2
2
π
r
Or
h
=
S
2
π
r
−
2
π
r
2
2
π
r
h
=
S
2
π
r
−
2
π
r
2
2
π
r
h
=
S
2
π
r
−
r
2
r
h
=
S
2
π
r
−
r