X+y^2=10
minus x from both sides
y^2=10-x
sub (10-x) for y^2 in other equation
Q=x(10-x)
Q=10x-x^2
now find the maximum value
take the derititive
dQ/dx=10-2x
it is zero at x=5
below that, it is positive
after 5, it is negative
max at x=5
solve for y
y^2=10-x
y^2=10-5
y^2=5
sqrt both sides
y=√5
x=5
y=√5
the max value is 25
If there is 2 field trips with 23 students and each student costs $5 dollars. So for the 1st trip it will be 23x$5
1st trip- $115
And on the second trip its the same because there is 23 students that cost 5 dollars so it will be 23x5
2nd trip- $115
To find out the total of both trips youd add 115+115/
In total, for both trips it will cost $230 dollars.
Y - y1 = m(x - x1)
slope(m) = -4/3
(0,-12)....x1 = 0 and y1 = -12
sub
y - (-12) = -4/3(x - 0) =
y + 12 = -4/3(x - 0) <=== point slope form
y + 12 = -4/3x
y = -4/3x - 12 <=== slope intercept form
y = -4/3x - 12
4/3x + y = -12
4x + 3y = -36 <=== standard form
3(x+5)=39
3 * x + 3 * 5 = 39
3x + 15 = 39
3x = 39 - 15
3x= 24
x = 8
For the first 60 positive integers, a = 1, n = 60, l = 60.
Sn = n/2(a + l)
s = 60/2(1 + 60) = 30(61)
For the next 60 positive integer, a = 61, n = 60, l = 120
Sum = 60/2(61 + 120) = 30(61 + 120) = 30(61) + 30(120) = s + 3600
Sum of first 120 positive integers = s + s + 3600 = 2s + 3600
