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)
1 gal + 2 qt + 3 pts
1 gal + 3 qt + 7 pts
1 gal + 3 qt + 3 pts
----------------------------add
3 gal + 8 qt + 13 pts
1 gal = 4 qts
1 gal = 8 pts
1 pt = 1/2 qt
13 pts = 6.5 qts
now we have 3 gal + (8 + 6.5) = 3 gal + 14.5 qts
12 qts = 3 gal
now we have : 6 gal + 2.5 qts or 6 gal + 2 qts + 1 pt <=== answer
Answer: y = −2
Step-by-step explanation:
- 0 = −4x − 2
- x = −1/2
- To find the y-intercept, substitute in 0 for x and solve for y. Y = − 4 * 0 −2
- Which then gives you y = - 2
* Hopefully this helps:) Mark me the brainliest:)
Answer:
<h2><em>
2(3s-14)</em></h2>
Step-by-step explanation:
Given the angles ∠ABF=8s-6, ∠ABE = 2(s + 11), we are to find the angle ∠EBF. The following expression is true for the three angles;
∠ABF = ∠ABE + ∠EBF
Substituting the given angles into the equation to get the unknown;
8s-6 = 2(s + 11)+ ∠EBF
open the parenthesis
8s-6 = 2s + 22+ ∠EBF
∠EBF = 8s-6-2s-22
collect the like terms
∠EBF = 8s-2s-22-6
∠EBF = 6s-28
factor out the common multiple
∠EBF = 2(3s-14)
<em></em>
<em>Hence the measure of angle ∠EBF is 2(3s-14)</em>
Answer:
5.85 m
Step-by-step explanation:
The width of the sand road can be calculated knowing its area and the dimensions of the rectangular garden as follows:
<u>Where:</u>
Ag: is the area of the rectangular garden
a: is the length of the rectangular garden = 50 cm = 0.5 m
b: is the width of the rectangular garden = 34 m
<u>Where</u>:
As: is the area of the sand road
The relation between the area of the sand road and the area of the rectangular garden is the following:
By solving the above equation for x we have two solutions:
x₁ = -23.10 m
x₂ = 5.85 m
Taking the positive value, we have that the width of the sand road is 5.85 m.
I hope it helps you!
