Line EF and Line Segment EF
this is probs wrong, but idk bc there aren't any answer choices
Answer:

Step-by-step explanation:
<h3>Derivative </h3>

Since, the derivative of e^x is e^x and e^(yx) is ye^(yx)
Step-by-step explanation:
f(x) = x² + x + 3/4
in general, such a quadratic function is defined as
f(x) = a×x² + b×x + c
the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 1
c = 3/4
x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =
= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =
= (-1 ± sqrt(2)i)/2
x1 = (-1 + sqrt(2)i) / 2
x2 = (-1 - sqrt(2)i) / 2
remember, i = sqrt(-1)
f(x) has no 0 results for x = real numbers.
for the solution we need to use imaginary numbers.
Answer:
12.8mph
Step-by-step explanation:
Let 'w' represent her walking speed
->if her jogging speed is 8mph faster than her walking speed
Jogging speed= w+8
Using the time equation i.e time = dist/speed
For walking when distance is 8miles
time = dist / speed= 8/w hrs
For jogging when distance is 13miles
time = dist / speed= 13/(w+8) hrs
Equating the time
jog time = walk time
13/(w+8) = 8/w
By cross multiplying it
13w = 8w + 64
5w = 64
w=64/5
w=12.8mph
Therefore, 12.8mph is her walking speed.
Your answer is C. Think of the x axis as the ground.