Answer:
5 + sqrt(65), 5 - sqrt(65)
Step-by-step explanation:
x + y = 10
x×y = -40
so, two equations with 2 variables. using one equation to express one variable by the second. and then using the second equation to solve and calculate the second variable. and then by knowing the second variable we calculate the first.
=> x = 10 - y
=> (10-y)×y = -40
=> -1×(10-y)×y = 40
=> (y-10)×y = 40
=> y^2 - 10×y - 40 = 0
remember the solution for a quadratic equation
a×x^2 + b×x + c
is
x = (-b ± sqrt(b^2 - 4×a×c)/(2×a)
in our case
a = 1
b = -10
c = -40
and we used "y".
so,
y = (--10 ± sqrt((-10)^2 - 4×1×(-40))/(2×1) =
= (10 ± sqrt(100 + 160))/2 = (10 ± sqrt(260))/2 =
= (10 ± sqrt (4×65))/2 = 10/2 ± 2×sqrt(65)/2 =
= 5 ± sqrt(65)
=>
for y = 5 + sqrt(65), x = 10 - 5 - sqrt(65) = 5 - sqrt(65)
for y = 5 - sqrt(65), x = 10 - 5 + sqrt(65) = 5 + sqrt(65)
Answer:
it's (1, 2)
Step-by-step explanation:
you just look at where the 2 lines cross
Answer:
There were 84 students in 6th grade, 68 in 7th grade, and 90 in 8th grade.
Step-by-step explanation:
Given that Elk middle school has 264 students, and there are 84 students in the eighth grade, and there are 22 fewer seventh-grade students than sixth-grade students, to determine how many students there are in each grade, the following calculation must be performed:
(264 - 84) / 2) - 22 = X
(180/2) - 22 = X
90 - 22 = X
68 = X
6th grade = 84
7th grade = 68
8th grade = 90
Therefore, there were 84 students in 6th grade, 68 in 7th grade, and 90 in 8th grade.
Answer:
Step-by-step explanation:
using sin formula
