The correct answer for this question would be option B) 18.8 units or the second option.
Consider the function

, which has derivative

.
The linear approximation of

for some value

within a neighborhood of

is given by

Let

. Then

can be estimated to be

![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since

for

, it follows that

must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function

. This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
Answer:
35/40
Step-by-step explanation:
7/8, with the given denominator 40
So how do we get 8, to turn into 40?
Simple multiplication.
8x5 = 40, and whatever we multiply times the denominator, we multiply for the numerator.
So
7x5 = 35
The answer should be 35/40
Answer: 3 or the whole backyard
Step-by-step explanation: In a hour she can rack 3 backyard. The question asked that how much of a backyard can she rake in 1 hour. She can rake the whole backyard in a hour.
Easiest way is if you substitute each point (x,y) into each set of equations and both points work for both equations in the system of equations, then it is the correct answer
Otherwise substitute one equation for y in the other equation:
2x + 6 = x^2 + 5x + 6
-2x - 6. -2x -6
0 = x^2 + 3x. Factor
0 = x (x + 3)
Solve: x = 0. x + 3 = 0. ——> x = -3. Substitute into one original equation to get y value for
y = 2x + 6.
y = 2(0) + 6. y = 2(-3) + 6
y = 6. y = -6 + 6 —-> y = 0
(0 , 6) And. (-3 , 0)