The answer is B.) F and I
The solution is shown in the graph attached.
Explanation:
I am goind to teach you how to get that solution.
1) Restricctions: both x and y cannot be negative, i.e.
x ≥ 0 and y ≥ 0⇒ the solution is on the
first quadrant.2) Using the prices of the tickets, $ 8 for adults, and $ 6 for children, the linear equation for the
costs is: cost = 8x + 6y.3) Since the radio station is willing to spend a
maximum of $ 172 you have the final restriction:
⇒ 8x + 6y ≤ 172.4) Then the solutions that meet the three restrictions (x ≥0, y ≥ 0, and 8x + 6y ≤ 172) is found graphically by drawing the line 8 x + 6y = 172
5) To draw the line 8x + 6y = 172, use the axis intercepts:
x = 0 ⇒ y = 172/6 = 86/3 ⇒ point (0, 86/3)
y = 0 ⇒ x = 172/8 = 86/4 ⇒ point (86/4, 0)
6) Once you have the line you
shade the region that is between the line and the two axis. That region contain of the possible solutions.
The answer is 20.
First you would multiple the negative with the negative 12, then you just add/subtract.
Answer:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
Step-by-step explanation:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)
Consider x^2-2/3x+1/3
Remember that (a-b)^2 =a^2-2ab+b^2
x^2=a^2
a=x
-2/3x= -2*x*b
b=1/3
S0 (x-1/3)^2= x^2-2/3x+1/9
x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
This is the same as the last question you asked. 63 + 61 = 124. 180 - 124 = 56. Angle ACB is 56 degrees.