<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
Answer:
-86
Step-by-step explanation:
-8((-11/4)-8)
((11/4-8)
8x(-43/4)
-2×43=-86
A=(v-u)/t
We know a=2.5, v=40, u=10, so:
2.5=(40-10)/t
2.5t=40-10
2.5t=30
Divide both sides by 2.5
t=12 It takes 12 seconds for the car to reach 40m/s
To find out how far it went, you do:
d=v+1/2 at^2
d=10+1/2(2.5)(12)^2
d=10+1/2(2.5)(144)
d=10+(1.25)(72)
d=10+90
d=100m
Hope this helps :)
Answer:
(d) f(x) = -x²
Step-by-step explanation:
For the vertex of the quadratic function to be at the origin, both the x-term and the constant must be zero. That is, the function must be of the form ...
f(x) = a(x -h)² +k . . . . . . . . . . vertex form; vertex at (h, k)
f(x) = a(x -0)² +0 = ax² . . . . . vertex at the origin, (h, k) = (0, 0)
Of the offered answer choices, the only one with a vertex at the origin is ...
f(x) = -x² . . . . . a=-1
