<span>X^3+6x^2-17x+2-x^3-x^2-11x+36 = </span>5x^2 - 28x + 38
(if 2 is a square)
6xy ( x2 - xy + y2)
= 6x3y - 6x2y2 - 6xy3 --------------the threes and twos on this line are squares and cubics. the six is just a whole number
D= # of dimes
q= # of quarters
QUANTITY EQUATION:
d + q= 64
COST EQUATION:
0.10d + 0.25q= $9.25
STEP 1:
multiply quantity equation by -0.10 to be able to eliminate the d term in step 2
(-0.10)(d + q)= (-0.10)(64)
-0.10d - 0.10q= -6.40
STEP 2:
add equation from step 1 to cost equation to eliminate the d term and solve for q
Add
0.10d + 0.25q= $9.25
-0.10d - 0.10q= -6.40
0.15q= 2.85
divide both sides by 0.15
q= 19 quarters
STEP 3:
substitute q value in step 2 into either original equation to find d value
d + q= 64
d + 19= 64
subtract 19 from both sides
d= 45 dimes
CHECK:
0.10d + 0.25q= $9.25
0.10(45) + 0.25(19)= 9.25
4.50 + 4.75= 9.25
9.25= 9.25
ANSWER: There are 45 dimes and 19 quarters.
Hope this helps! :)
The probability of getting a red card is 1/2 and the probability of getting a 10 is 4/52. To find the probability of getting both a red card and a 10, multiply the two.
(1/2)(4/52) = 4/104
4/104 = 1/26
Therefore, the probability is 1/26
