May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
X axis : -t1 cos 60 + t2 cos 60 = 0
t2 cos 60 = t1 cos 60 ; t2 = t1
Y axis : t1 sin 60 + t2 sin 60 - 150 = 0
since t2 = t1
2t1 sin 60 = 150
t1 ( sqrt(3)/2) = 150/2
t1 = 50 [sqrt(3)]
t2 = 50 [sqrt(3)]
t3 = 150 N
hope this helps
Answer:
73
Step-by-step explanation:
...wait 69?!?!?!?!
Answer:
528
Step-by-step explanation:
So naturally at first sight for this question we would think -->
Oh 3 consecutive integers = 66 --> 66/3 = 22 (n-1), (n+1) so --> 21, 22, and 23.
But no. At second look it is the sum of 3 consecutive integers is greater than 66. So we find the next possible pair since it says smallest possible product.
We get the set (22, 23, 24). => Multiply the least and greatest integers together respectively 22 and 24 which amounts to => 528
And thus, we have out answer of 528
Hope this helps!
Answer:
3x + 2y =18
Y= -2/3x + 12
3x + 2(-2/3x+12) = 18
3x + (-4/3x+24) = 18
3x + -1 1/3x +24 =18
1 2/3x +24 = 18
1 2/3x = -6
5/3x = -6
x = -3.75
Step-by-step explanation:
