The diameter is 57 cm what is the radius and area write your solution below
2 answers:
You might be interested in
Note that the notation 
is just another way of writing 
. Here, we're simply taking the expression for f(x) and adding the expression for g(x) to it. Together, we have:
Which means that
And from there, it's just a matter of combining like terms to simplify.
Which means that
And from there, it's just a matter of combining like terms to simplify.
Answer:
Step-by-step explanation:
we have
50%, 2/25, 5, and 0.31
1) 50%
we know that
2) 2/25
Multiply by 4/4
3) 5
Multiply by 100/100
4) 0.31
Multiply by 100/100
Now all numbers have the same denominator
so
therefore
The parking lot is located along the front and one side of the school
building.
- The distance <em>x</em> with which the lot should be expanded is <u>60 feet</u>.
Reasons:
The given dimensions in the compound are;
Length of the school and the old parking lot = 300 + 75
Width of the school and the old parking lot = 165 + 75
Length of the school, the old and new parking lot = 300 + 75 + x
Width of the school and the old and new parking lot = 165 + 75 + x
Area of the new parking lot the school wants = Double the old parking lot
In terms of areas, we have;
New compound - School = 2 × (old compound - school)
Which gives;
(300 + 75 + x) × (165 + 75 + x) - 300 × 165 = 2 × ((300 + 75) × (165 + 75) - 300 × 165)
2 × (375 × 240 - 49,500) = (375 + x) × (240 + x) - 49,500
Simplifying gives;
81,000 = x² + 615·x + 40,500
x² + 615·x + 40,500 - 81,000 = 0
x² + 615·x - 40,500 = 0
Factorizing with a graphing calculator gives;
(x - 60)·(x + 675) = 0
Therefore;
x = 60 or x = -675
Given that <em>x</em> is a natural number, we have, the correct value is; x = 60;
The distance with which the lot should be expanded, x = <u>60 feet</u>.
Learn more about writing equations here:
