You use the quadratic formula:
2x {}^{2} - 3x = 5
2x {}^{2} - 3x - 5 = 0
x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2}
x = \frac{3 + - \sqrt{49} }{4}
x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4}
x = \frac{5}{2} \: and \: x = - 1
Answer:
r u trying to look for the values of D and H?
D (3)(-9) = 0
D -27 = 0
D = 27
H - 24 = 4
H = 4 + 24
H = 28
Let number of fives = x then number of ones = 3x and number of tens = x - 1
so we can create the equation
x + 3x + x-1 + y = 26 where y = number of twenties
so
5x + y = 27
also we have the equation
5x + 3x + 10(x - 1) + 20y = 120
18x + 20y = 130..................................(1)
5x + y = 27 multiply by -20:-
-100x - 20y = -540..............................(2)
Adding equation (1) and (2)
-82x = -410
x = 5, that is 5 fives
Now plug x = 5 into equation 1:-
18(5) + 20y = 130
20y = 40
y = 2 , that is 2 twenties
So the answer is there are (3x) = 15 ones , 5 fives, 4 tens and 2 twenties
Answer:
x=11,10,9,8,7,6,5,4,3,2,1,0, and all the negative numbers could be subsituted for x
Step-by-step explanation:
You will use a2+b2=c2 (squared).The leg will represent the a2 or either b2 and the hypotenuse will be the c2.After filling in the variables you will then subtract the a2 which is the leg from the c2 which is the hypotenuse.And then you will factor it or square root it.If you square root it and it’s not a feasible answer then use the Christmas tree or birthday cake method to factor the answer and then you’ll have the value of b2 which is the unknown leg.