I must confess that I was about to pass this question by, but I was captured by the respectful and dignified way in which you asked for help.
A careful reading of the problem gives you two equations in two unknowns, which you can then solve as simultaneous equations. Here's how it looks:
Call 'C' the price of the senior <u>C</u>itizen ticket.
Call 'S' the price of the <u>S</u>tudent ticket.
On the first night . . . 10 C + 12 S = 208
On the second night . . . 8C + 3 S = 74
Those are your two simultaneous equations. Now the idea is to multiply or divide each side of one equation in such a way that when you add or subtract it from the other equation, one of the variables will become a zero quantity ... you'll be left with an equation in one variable, which you can easily solve. THEN, knowing the value of one variable, you can put it back into one of the original equations,and find the value of the other variable.
This all sounds more complicated than it is. Here's how it goes:
We have . . .
10 C + 12 S = 208 and
8C + 3 S = 74
I'm going to multiply each side of the second equation by 4, and then write it under the first one:
10 C + 12 S = 208
32 C + 12 S = 296
Now, subtract the lower equation from the upper one, and you get . . .
- 22 C + 0 = - 88
Divide each side of this one by -22 and you have <em>C = $4.00</em> .
THAT's what you need, to blow the whole problem wide open. Knowing
the value of 'C', let's substitute it into the equation for the first night:
10 C + 12 S = 208
10(4) + 12 S = 208
40 + 12 S = 208
Subtract 40 from each side : 12 S = 168
Divide each side by 12 : <em>S =</em><em> $ 14.00 </em>.
Finally, as we look over our results, and see that Students have to pay $14 to see the show but Seniors can get in for only $4 , we reflect on this ... or at least I do ... and realize that getting old is not necessarily all bad.
Answer:
Step-by-step explanation:
B(2,10); D(6,2)
Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)
Rhombus all sides are equal.
AB = BC = CD =AD
distance = √(x2-x1)² + (y2- y1)²
As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x
A(X,0)
AB = AD
√(2-x)² + (10-0)² = √(6-x)² + (2-0)²
√(2-x)² + (10)² = √(6-x)² + (2)²
√x² -4x +4 + 100 = √x²-12x+36 + 4
√x² -4x + 104 = √x²-12x+40
square both sides,
x² -4x + 104 = x²-12x+40
x² -4x - x²+ 12x = 40 - 104
8x = -64
x = -64/8
x = -8
A(-8,0)
Let C(a,b)
M is AC midpoint
(-8+a/2, 0 + b/2) = M(4,6)
(-8+a/2, b/2) = M(4,6)
Comparing;
-8+a/2 = 4 ; b/2 = 6
-8+a = 4*2 ; b = 6*2
-8+a = 8 ; b = 12
a = 8 +8
a = 16
Hence, C(16,12)
First of all, Rodger must be making some serious cash to have 47 cars. Going back to the question, he can not group the cars in more than 2 ways. He can only group it in 1 way, 47 groups of 1. Pls brainliest
Answer:
A0.15
Step-by-step explanation:
