Answer:
There are two pairs of solutions: (2,7) and (-1,4)
Step-by-step explanation:
We will use substitution.
y = x^2 + 3
y = x +5
Since the second equation is equal to y, replace y in the first equation with the second equation.
y = x^2 + 3
x + 5 = x^2 + 3
Rearrange so that one side is equal to 0.
5 - 3 = x^2 - x
2 = x^2 - x
0 = x^2 - x - 2
You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).
a = 1, b = -1, c = -2
Zeros = 
and 
Zeros = 2 and -1
Now that you have your x values, plug them into the equations to find their corresponding y values.
y = x^2 + 3
y = (2)^2 + 3
y = 7
Pair #1: (2,7)
y = x^2 + 3
y = (-1)^2 + 3
y = 4
Pair #2: (-1,4)
Therefore, there are two pairs of solutions: (2,7) and (-1,4).
You can make 3 and 3/7 into a improper fraction as 24/7 by multiplying 7 by 3 which is 21 and then adding that to 3 which gives you to answer 24/7. To find the reciprocal you simply flip the fraction which gives you 7/24
Answer:
B. 2
Step-by-step explanation:
Based on the Midsegment of a trapezoid theorem:
GH = (BC + FE)/2
9x - 3 = (19 + (5x + 1))/2) (substitution)
9x - 3 = (20 + 5x)/2
Cross multiply
2(9x - 3) = 20 + 5x
18x - 6 = 20 + 5x
Collect like terms
18x - 5x = 20 + 6
13x = 26
Divide both sides by 13
x = 26/13
x = 2
5Red + 6Blue + 4 Black
Total = 5 + 6 = 4 = 15
Probality of Blue, Blue.
P(Blue) = 6/15 = 2/5
P(Blue, Blue) = (2/5)*(2/5) = 4/25
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)
