Answer:
The area of the region between the two curves by integration over the x-axis is 9.9 square units.
Step-by-step explanation:
This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:
Given that resulting expression is a second order polynomial of the form 
, there are two real and distinct solutions. Roots of the expression are:
and 
.
Now, it is also required to determine which part of the interval 
is equal to a number greater than zero (positive). That is:
and 
.
Therefore, exists two sub-intervals: ![[-5, -4.899]](https://tex.z-dn.net/?f=%5B-5%2C%20-4.899%5D)
and ![\left[4.899,5\right]](https://tex.z-dn.net/?f=%5Cleft%5B4.899%2C5%5Cright%5D)
. Besides, 
in each sub-interval. The definite integral of the region between the two curves over the x-axis is:
![A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx](https://tex.z-dn.net/?f=A%20%3D%20%5Cint%5Climits%5E%7B-4.899%7D_%7B-5%7D%20%5B%7B1%20-%20%28x%5E%7B2%7D-24%29%5D%7D%20%5C%2C%20dx%20%2B%20%5Cint%5Climits%5E%7B4.899%7D_%7B-4.899%7D%20%5C%2C%20dx%20%2B%20%5Cint%5Climits%5E%7B5%7D_%7B4.899%7D%20%5B%7B1%20-%20%28x%5E%7B2%7D-24%29%5D%7D%20%5C%2C%20dx)
The area of the region between the two curves by integration over the x-axis is 9.9 square units.
As stated 1 mile equals to 1.6 km,
so do calculation to get the 6 km
1 mile = 1.6 km
Then try to divide the 6 km with 1 mile value in km which is 1.6 km so that we know exactly how many miles required to get 6 km
So the answer is 3.75 miles
To prove this
If 1 mile = 1.6,
Then,
3 miles = (1.6 x 3 ) which produce 4.8 km
0.75 mile = ( 0.75 x 1.6 ) equals to 1.2 km
4.8km + 1.2km = 6km , which proves that
3 miles + 0.75 mile = 6km
Answer:
Step-by-step explanation:
i Don't now how to explain the answer
Answer:
65625/4(x^5)(y²)
Step-by-step explanation:
Using binomial expansion
Formula: (n k) (a^k)(b ^(n-k))
Where (n k) represents n combination of k (nCk)
From the question k = 5 (i.e. 5th term)
n = 7 (power of expression)
a = 5x
b = -y/2
....................
Solving nCk
n = 7
k = 5
nCk = 7C5
= 7!/(5!2!) ------ Expand Expression
=7 * 6 * 5! /(5! * 2*1)
= 7*6/2
= 21 ------
.........................
Solving (a^k) (b^(n-k))
a = 5x
b = -y/2
k = 5
n = 7
Substituting these values in the expression
(5x)^5 * (-y/2)^(7-5)
= (3125x^5) * (-y/2)²
= 3125x^5 * y²/4
= (3125x^5)(y²)/4
------------------------------------
Multiplying the two expression above
21 * (3125x^5)(y²)/4
= 65625/4(x^5)(y²)
