Answer: n=-36
Step-by-step explanation: Hope this help :D
Use the distributive property to express the sum of the 2 whole numbers 15 and 30 with common factors as a multiple of two whole numbers with a sum of no common factor the solution would be (15 plus 30) = 15(1+2)
Answer:
Width = 52cm
Step-by-step explanation:
Step one:
given data
Area= 3276 cm squared
lenght= 63cm
Required
The width
Step two:
We know that
Area= L*W
substitute
3276= 63*W
divide both sides by 63
W= 3276/63
W=52cm
First of all we need to find a representation of C, so this is shown in the figure below.
So the integral we need to compute is this:

So, as shown in the figure, C = C1 + C2, so:
Computing first integral:
Applying derivative:

Substituting this value into

Computing second integral:
Applying derivative:

Substituting this differential into


We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:
![I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}](https://tex.z-dn.net/?f=I_%7B2%7D%3D%20-8%5Cint_%7B4%7D%5E%7B8%7D%7Ddx%3D-8%5Bx%5D%5Cright%7C_4%20%5E%7B8%7D%3D-8%288-4%29%20%5Crightarrow%20%5Cboxed%7BI_%7B2%7D%3D-32%7D)
Finally: