13. Pick a point and see which formula works.
Ay = -4, A'y = 7. Only the formula of selection D makes that translation.
14. Use the compound interest formula A = P*(1 +r/n)^(nt).
..1500*1.015^80 = 4935.99, matching selection C
15. The lid has a perimeter of 90", so the area of the sides is
.. 90" * 24" = 2160 in^2
The area of the lid is
.. 30" * 15" = 450 in^2
The gray area is (2160 -450) in^2 = 1710 in^2 larger, corresponding to selection C.
16. The only formula that maps (7, -1) to (21, -3) is that of selection D.
_____
The middle two problems are the only ones that require you to have prior knowledge. The others could be answered simply by seeing if the answers work.
The pass rusher, which is typically line up on the line of scrimmage.
Have a Great Day :)
Answer:
D
Step-by-step explanation:
For simplify the work we can start to factorise all the possibles expressions:
2x + 8.
8 is multiple of 2, so it can became
2(x+4)
x^2 - 16 this is a difference of two squares, so it can be rewritten as:
(x+4)(x-4)
x^2 + 8x + 16
we have to find two numbers whose sum is 8 and whose product is 16
the two number are 4 and 4
it becames:
(x+4)(x+4)
x+ 4 can‘t be simplified
if we look at the expression, we can find that x-4 appears at the numerator so
x^2 - 16 must be at numerator
but the second factor (x+4) doesn’t appear, so has been simplified. This situation can be possible only in the D option
in fact
(x+4)(x-4)/2(x+4) * (x+4)/(x+4)(x+4)
it became
(x+4)(x-4)/2 * 1/(x+4)(x+4)
(x-4)/2(x+4)
Answer:
only -2 is in the domain
Step-by-step explanation:
For negative values under the sqrt, there is no real answer, hence
-2x ≥ 0
only for x=-2 this is the case.
The domain represents all the valid input values for x.
For each curve, plug in the given point 
and check if the equality holds. For example:
(I) (2, 3) does lie on 
since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.
For part (a), compute the derivative 
, and evaluate it for the given point . This is the slope of the tangent line at the point. For example:
(I) The derivative is
so the slope of the tangent at (2, 3) is
and its equation is then
For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, 
. For example:
(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation
