so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.
in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.
now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B7%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B%5Cfrac%7B19%7D%7B2%7D%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D-7%20%5Cright%29%5E2%2B%5Cleft%28%20%5Cfrac%7B7%7D%7B2%7D-4%20%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B5%7D%7B2%7D%5Cright%29%5E2%2B%5Cleft%28%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Cboxed%7BAM%5Capprox%202.549509756796392%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
explanation:
The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.
An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:
x=(x1+x2)/2
y=(y1+y2)/2
Solving each equation for (x2,y2) yields the solution (0,-14).
please mark me as brainliest
You need to find the amount subject to withholding, subtracting from the weekly salary the amount for one withholding allowance for weekly salaries, which is 77.90$:
830 - 77.90 = 752.1 $.
Then, look in the Fed Tax tables (
http://www.opers.ok.gov/Websites/opers/images/pdfs/2016-Fed-Tax-Tables.pdf ) for a married person with a weekly payroll.
You previously found an amount of 752.1 which is greater than 521 but less than 1613$: therefore the income tax to withhold is 35.70$ + 15% of excess over $521.
Therefore, calculate the income tax due: 35.70 + (752.1 - 521) × 15 ÷ 100 = 70.37$
The total amount of income tax that will be withheld is 70.37$
The volume of a rectangular prism is given by the multiplication of its three dimensions: if 
are the width, height and length of the prism, the volume is
So, in your case, you have
Answer: 77.05
Step-by-step explanation: 1) Draw a right triangle with these measures:
angle with the horizontal: 6 degress
horizontal segment: 5
height: x
2) Definition of tangent : tan(6°) = opposite leg / adjacent leg = x / 5 =>
3) Solve for
x = 5tan(6°) = 77.05
