Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)
Answer:
Search this up on google to find out what the answer is because no one is going to know how to do it.
X + x + 10 + x + 20 + x + 30 = 84.
4x + 60 = 84.
84 − 60 = 24
4x = 24.
24 / 4 = 6 = x.
Answer:
Day One: 6 fish
Day Two: 16 fish
Day Three: 26 fish
Day Four: 36 fish
Domain: f (x) is the set of all values for which the function is defined
range: the set of all values that f takes
hope this helps
Step-by-step explanation:
The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.
for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.
The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.
substituting 36 for x and 10 for y in eq 1;
3(36) + 10 = 108 + 10 = 118 total students for Green High School
substituting 36 for x and 10 for y in eq2;
4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school
