Answer:
(10^2)*2, 10*3, 10/2
Step-by-step explanation:
Thats my best guess to your very vague question
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...
I would think it would be B
We have that
<span>question 1
Add or subtract.
4m2 − 10m3 − 3m2 + 20m3
=(4m2-3m2)+(20m3-10m3)
=m2+10m3
the answer is the option
</span><span>B: m2 + 10m3
</span><span>Question 2:
Subtract. (9a3 + 6a2 − a) − (a3 + 6a − 3)
=(9a3-a3)+(6a2)+(-a-6a)+(-3)
=8a3+6a2-7a-3
the answer is the option
</span><span>B: 8a3 + 6a2 − 7a + 3
</span><span>Question 3:
A company distributes its product by train and by truck. The cost of distributing by train can be modeled as −0.06x2 + 35x − 135, and the cost of distributing by truck can be modeled as −0.03x2 + 29x − 165, where x is the number of tons of product distributed. Write a polynomial that represents the difference between the cost of distributing by train and the cost of distributing by truck.
we have that
[</span>the cost of distributing by train]-[the cost of distributing by truck]
=[−0.06x2 + 35x − 135]-[−0.03x2 + 29x − 165]
<span>=(-0.06x2+0.03x2)+(35x-29x)+(-135+165)
=-0.03x2+6x+30
the answer is the option
</span><span>C: −0.03x2 + 6x + 30
</span><span>
</span>
1 gallon = 16.8 miles
265 gallons = 16.8 x 265 = 4452 miles
Answer: 4452 miles
