For
, we have
So for
to be continuous at
, we require that the limit as
is equal to 4.
51 ft is the answer
Explanation
The hight of the man divided by the hight of school ( which is unknown )
Equal
The length of the shadow of the man divided by the shadow of the school
Cross multiplication and find the value of unknown x which is 51... and ya
Answer:
there is no greatest load
Step-by-step explanation:
Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:
x ≥ y +600 . . . . . my truck can carry at least 600 pounds more
x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers
Combining these two inequalities, we have ...
4/3y ≥ x ≥ y +600
1/3y ≥ 600 . . . . . . . subtract y
y ≥ 1800 . . . . . . . . multiply by 3
My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.
By exterior angle of triangle:
17x+4+8x+1=80
x=3
First you Change the mixed number to an improper fraction (14/3) then you simplify (1/4 X 7/3) and then you multiply 1/4 X 7/3 = 7/12 :)
