Hi there!
a. 8x^2 - 1
(x^2 - 4x + 5) + (7x^2 + 4x - 6)
= x^2 + 7x^2 - 4x + 4x - 1
= 8x^2 - 1
b. 5x^2 + 7x - 7
(7x^2 + 4x - 6) - (2x^2 - 3x + 1)
= 7x^2 + 4x - 6 - 2x^2 + 3x - 1
= 7x^2 - 2x^2 + 4x + 3x - 6 - 1
= 5x^2 + 7x -7
Hope this helps!
That does look impossible someone plz help this man out or girl I don’t wanna assume genders ya know
9514 1404 393
Answer:
x ≠ 3
Step-by-step explanation:
In any case, the domain is restricted to values of the variable for which the function is defined. The value 1/0 is not defined, so the variable cannot allow the denominator to be zero. The denominator x-3 will be zero for x=3, so that value of the variable cannot be in the domain.
The domain is all real numbers except x=3.
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<em>Additional comments</em>
It is useful to become familiar with the domains of different functions. As we saw above, the reciprocal of 0 is undefined. The square root of a negative number is undefined. The log of a non-positive number is undefined. Trig functions are defined everywhere, but their inverse functions are not. Polynomial functions are defined everywhere, but ratios of polynomials have the same restriction on denominators that we see above.
1a) f(x) = I x+2 I. This is a piece-wise graph ( V form)
x = 0 →f(x) =2 (intercept y-axis)
x = -2→f(x) = 0 (intercept x-axis)
x = -3→f(x) = 1 (don't forget this is in absolute numbers)
x = -4→f(x) = 2 (don't forget this is in absolute numbers)
Now you can graph the V graph
1b) Translation: x to shift (-3) units and y remains the same, then
f(x-3) = I x - 3 + 2 I = I x-1 I
the V graph will shift one unit to the right, keeping the same y. Proof:
f(x) = I x-1 I . Intercept x-axis when I x-1 I = 0, so x= 1
The volume of a cone is (1/3) that of its similar cylinder.
Thus the volume of the cone B would be:
