Distributive property: a(b + c) = ab + ac
4n - 2 - 2n = 2(2n - 1 - n)
4n - 2 - 2n = 2n - 2 = 2(n - 1)
A:
Table 1:
f(x) = 8x
X= cups so
ounces =
8* # of cups
Table 2:
f(x) =16x
X= pints so
ounces =
16*# of pints
Table 3:
f(x)= 32x
X= quarts so
ounces=
32*# of quarts
Note: another way instead of f(x)=kx is y=kx
B:
Depending on the slope we can easily find which is the steepest slope by easily comparing the slopes and seeing which is greater.
T1: 8
T2: 16
T3: 32
So, looking at our slopes, table 3 has the greatest slope because it has the greater slope out of the other two.
C:
The phrase, "rate of change" is another way to say the slope. To list again,
T1: 8
T2: 16
T3: 32
The table with the smallest slope would be table 1 since it is less than the other two tables.
I hope that helps, if you need any more assistance or you have any questions, please feel free to ask!!
Answer: 4x³+29x²+40x-48
Step-by-step explanation:
To solve f(x)·g(x), you multiply them together.
(4x²+13x-12)(x+4) [distribute by FOIL]
4x³+16x²+13x²+52x-12x-48 [combine like terms]
4x³+29x²+40x-48
Now, we know that f(x)·g(x)=4x³+29x²+40x-48.
The second option. Angles 2 and 6 are corresponding angles, and 6 and 7 make a linear pair, so 2 and 7 are supplementary.
The first option doesn’t even mention angle 7, the third option doesn’t mention angle 2, and the last option doesn’t mention 7 either.
<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
c^2=81
c^2=a^2+b^2
b^2=c^2-a^2=81-49=32
Equation of given hyperbola:
..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given hyperbola:
center: (0,0)
a=3 (distance from center to vertices)
a^2=9
c=6 (distance from center to vertices)
c^2=36 a^2+b^2
b^2=c^2-a^2=36-9=25
Equation of given hyperbola:
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