PLEASE PLEASE PLEASEE PLEASEEEEESE HELPME!?!?!? IM SO BAD AT MATH PLEASE HELP!!! Find the average rate of change from x = 7 to x = 14 for the function f(x) = 0.01(2)x and select the correct answer below.
Select one:
a. 23.22
b. 35.84
c. 128
d. 163.84
IK THE ANSWER IS NOT D. 163.84
Answers
By definition, the average rate of change is given by:
We evaluate the function for the given values:
For x = 7:
For x = 14:
Then, replacing values we have:
Answer:
the average rate of change from x = 7 to x = 14 is:
a. 23.22
Related Questions
203.530 divided by 0.7=
What is the quotient of 75,120 ÷ 16?
The length of a square is 16m, what is the breadth of the square
What is 6,720 divided by 14??
Answers
Answer:
A) 2
Step-by-step explanation:
(0,3) and (2,7)
Use slope formula
(7-3)/(2-0) = 4/2 = 2
Answer:
A) 2
Step-by-step explanation:
A. True
B. False
Answers
answer TRUE because they would make a triangle
Answers
Answer:
area = 1500× 750 =
Step-by-step explanation:
we know area of rectangle
for length = l m
and width = b m
and perimeter
but one side length measures is not required because of the river so
He does not use the fence along the side of the river
so we use this formula
Perimeter = P = L + 2 b
Perimeter is 3000 m
so area will be
it is a quadratic function whose max or min will
occur at the average of the Solutions.
on Solving (3000 - 2b)b = 0
3000 - 2b = 0 or b=0
2b =3000
or
The average of the values are
so for max area we use b=
The Length is then L=3000 - 2(750) = 3000 - 1500 = 1500
for max area
length = 1500 m
bredth = 750 m
area = 1500× 750 =
Final answer:
The largest area that can be enclosed by Farmer Ed with 3000 meters of fencing along a river (with only three sides fenced) equals 1,125,000 square meters by using principles of mathematical optimization.
Explanation:
In this question, Farmer Ed wants to maximize the area of a rectangle with only three sides fenced, since one side borders on a river. We can use the principles of optimization in mathematics to solve this problem.
With 3000 meters of fencing for three sides, if we denote one side perpendicular to the river as X and the side parallel to the river (which forms the base of the rectangle) as Y, then, the perimeter would be Y+2X which is equal to 3000 meters. So, Y = 3000-2X.
The area A of a rectangle is length times width, or, in this case, A = XY. Substituting Y from the equation above: A = X(3000-2X) = 3000X - 2X^2. To maximize this area, we need to find values of X for which this equation has its maximum value.
The maximum or minimum of a function can be found at points where its derivative is zero. So, we take the derivative of A with respect to X, set it equal to zero, and solve for X.
The derivative, dA/dX is 3000 - 4X. Setting this equal to 0 gives X = 3000/4 = 750. So, the maximum area that Farmer Ed can enclose is when X is 750, and Y is 3000 - 2X = 1500, so the maximum area is 750 * 1500 = 1,125,000 square meters.
Learn more about Mathematical Optimization here:
#SPJ3
a. 51
b. 53
c. 54
d. 55
Answers
Final answer:
The missing number in the sequence is 54.
Explanation:
The missing number in the sequence is 54.
To identify the missing number, we need to observe the pattern in the sequence. The sequence decreases by 12 each time. Starting from 78, we subtract 12 to get 66, then subtract 12 again to get 54, and so on. Therefore, the missing number is 54.
Learn more about Finding the missing number in a sequence
B. All real numbers
C. All real numbers between -6 and -1 & -1 and 2
D. All real numbers between -8 and 4
Answers
I think that it's D because the y-values go from 4 and end at -8
cos A /(1- sin A) = (1 + sin A)/cos A
Answers
Answer:
answer is in exaplation
Step-by-step explanation:
cosA
+
cosA
1+sinA
=2secA
Step-by-step explanation:
\begin{lgathered}LHS = \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=\frac{cos^{2}A+(1+sinA)^{2}}{(1+sinA)cosA}\\=\frac{cos^{2}A+1^{2}+sin^{2}A+2sinA}{(1+sinA)cosA}\\=\frac{(cos^{2}A+sin^{2}A)+1+2sinA}{(1+sinA)cosA}\\=\frac{1+1+2sinA}{(1+sinA)cosA}\end{lgathered}
LHS=
1+sinA
cosA
+
cosA
1+sinA
=
(1+sinA)cosA
cos
2
A+(1+sinA)
2
=
(1+sinA)cosA
cos
2
A+1
2
+sin
2
A+2sinA
=
(1+sinA)cosA
(cos
2
A+sin
2
A)+1+2sinA
=
(1+sinA)cosA
1+1+2sinA
/* By Trigonometric identity:
cos² A+ sin² A = 1 */
\begin{lgathered}=\frac{2+2sinA}{(1+sinA)cosA}\\=\frac{2(1+sinA)}{(1+sinA)cosA}\\\end{lgathered}
=
(1+sinA)cosA
2+2sinA
=
(1+sinA)cosA
2(1+sinA)
After cancellation,we get
\begin{lgathered}= \frac{2}{cosA}\\=2secA\\=RHS\end{lgathered}
=
cosA
2
=2secA
=RHS
Therefore,
\begin{lgathered}\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA\end{lgathered}
1+sinA
cosA
+
cosA
1+sinA
=2secA