The semi-circles in the triangles represent the verteces or different angles on certain points of the triangle.
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Answer: Choice B
(-2, 5)
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Explanation:
The original system is
Multiply both sides of the second equation by 3. Doing so leads to this updated system of equations
Now add straight down
The x terms add to -4x+3x = -1x = -x
The y terms add to 3y+(-3y) = 0y = 0
The terms on the right hand sides add to 23+(-21) = 2
We end up with the equation -x = 2 which solves to x = -2
Now use this to find y. You can pick any equation with x,y in it
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-4x+3y = 23
-4(-2)+3y = 23
8+3y = 23
3y = 23-8
3y = 15
y = 15/3
y = 5
Or
x-y = -7
-2-y = -7
-y = -7+2
y = -5
y = 5
Either way, we get the same y value.
So that's why the solution is (x,y) = (-2, 5)
We will get the number of possible selections, and then subtract the number less than 25 cents.
We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.
That's 5*4*3 = 60 selections
Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels.
To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2 on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).
If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.
And there is 1 other selection less than 25 cents, 2 dimes and no nickels.
So that's 4+3+1 = 8 selections which we must subtract from the 60.
Answer 60-8 = 52 selections of coins worth 25 cents or more.
B=1/2
9/16=1/2b(3/4)
Divide both sides by 3/4
1/4=1\2b
Multiply both sides by 2
AM=3x+3
AB=8x-6
AB=2AM
AM=AB/2
AM=(8x-6)/2
AM=4x-3
3x+3=4x-3
4x-3x=3+3
x=6
AM=3x+3
AM=3(6)+3
AM=18+3
AM=21
