Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Triangle HAM looks like an isosceles triangle, and we know that in an isosceles triangle 2 of the angles are the same.
so :
180 - 106 = 74 ( sum of the other 2 angles )
74/2 = 37 ( sum of angle HMA / HAM )
angles on a straight line add up to 180.
180-37=143 ( angle x )
since triangle YMH is also an isosceles,
180-37=143
143/2=71.5 ( angle w )
82 Europe dduuxhxxhhchcch
So domain is the number you can use
range is the output your get from inputting the domain given
so from 2≤x≤5
since it is linear, we can be sure that we only need to test the endpoints of the domain to find the endpoints of the range
sub 2 for x
y=2(2)+1
y=4+1
y=5
sub 5 for x
y=2(5)+1
y=10+1
y=11
so range is from 5 to 11
in interval notation: [5,11]
in other notation 5≤y≤11
or
R={y|5≤y≤11}
X = the number of months the tire was used
y = the thickness of the tire tread
y = -(3/7)x + 20
Substitute x for 20 (the number of months the tire was used) and solve for y.
The next question: Substitute x for 49 (the number of months the tire was used) and solve for y.
