Answer:
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
Step-by-step explanation: See Annex
Green Theorem establishes:
∫C ( Mdx + Ndy ) = ∫∫R ( δN/dx - δM/dy ) dA
Then
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy
Here
M = 2x + cosy² δM/dy = 1
N = y + e√x δN/dx = 2
δN/dx - δM/dy = 2 - 1 = 1
∫∫(R) dxdy ∫∫ dxdy
Now integration limits ( see Annex)
dy is from x = y² then y = √x to y = x² and for dx
dx is from 0 to 1 then
∫ dy = y | √x ; x² ∫dy = x² - √x
And
∫₀¹ ( x² - √x ) dx = x³/3 - 2/3 √x |₀¹ = 1/3 - 0
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
Since the length is 15, and the width would be 15, you need to first do 15 x 15.
15 x 15 = 225
Multiply the 6 faces.
225x6
Your final answer would be 1,350.
Answer:
yes, i think it is
Step-by-step explanation:
Answer:
Find the midsegment of the triangle which is parallel to CA.
Tip
- A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle.
- This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side.
- If two segments are congruent, then they have the same length or measure.In other words, congruent sides of a triangle have the same length.
We have to find the segment which is parallel to CA.
From the given data,
The segment EG is the midsegment of the triangle
ABC.
So we have,
A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side.
~
let
(hypotenuse ) c = 10 in
(perpendicular) a = 8 in
(base) b =?
By using Pythagoras theorem
c^2= a^2+b^2
(10)^2 = (8)^2 +(b)^2
100 = 64 +(b)^2
100- 64 = (b)^2
36 = (b)^2
√36 = b
6 = b
hence b= 6 in
