Answer:
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line
is
Step-by-step explanation:
Given:
To Find:
Equation of line passing through ( 16, -7) and is perpendicular to the line
Solution:
...........Given

Comparing with,
Where m =slope
We get
We know that for Perpendicular lines have product slopes = -1.

Substituting m1 we get m2 as

Therefore the slope of the required line passing through (16 , -7) will have the slope,
Now the equation of line in slope point form given by
Substituting the point (16 , -7) and slope m2 we will get the required equation of the line,
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line
is
Answer:

Step-by-step explanation:
We want to find an expression that has a value of

First option:

Second option:

Correct option
Third option

Fourth option:

Well the equation is (A^2)+(B^2)=(C^2)
A= x
B= 2x
C will always be the longest side because it is the Hypotenuse = 25
So if you plug in those numbers into the equation...
(x^2) + (2x^2) = (25^2)
x^2 + 4x^2 = 625
Combine like terms
5x^2 = 625
Divide by five to both sides
x^2 = 125
Then Square root,
x = sqrt(125)
x = sqrt(25* 5)
x = 5sqrt(5)
Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

where
. Each interval has length
.
At these sampling points, the function takes on values of

We approximate the integral with the Riemann sum:

Recall that

so that the sum reduces to

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

Just to check:

C.35cm
Should be the right answer