Answer:
The width of the deck is 2m .
Step-by-step explanation:
Let the width of the pool be represented by w. As stated in the problem a deck of uniform width is to be built so the total width of the pool and deck will be w+w = 2w.
Now the total dimensions can be taken as
10m+ 2w and 5m + 2w of both the pool and deck.
Area = length * breadth
126m ²= (10m+ 2w) ( 5m + 2w )
126= 50 + 30 w+ 4w²
0= 50 -126+ 30 w+ 4w²
0 = 30 w+ 4w²- 76
Taking 2 as common
0= 2( 15 w+ 2w² - 38)
2w²+ 15 w - 38= 0
The above equation is the quadratic equation and can be solved as follows.
a= 2 , b= 15 and c= -38
b= -b±√b²- 4ac/2a
Putting the values
b= - 15±√15²- 4( 2)(-38)/2(2)
b= - 15± 23/4
b= - 38/4 or 8/4
b= 2 m
The width cannot be negative so we ignore the negative value
The width of the deck is 2m .
The can be checked by putting the value in the original equation.
126m ²= (10m+ 2w) ( 5m + 2w )
126m ²= (10m+ 2(2)) ( 5m + 2(2) )
126m ²= (10m+ 4) ( 5m + 4 )
126m ²= 14*9
126m ²= 126m ²
Answer:
±3
Step-by-step explanation:
4 - 7 = -3
10 - 7 = 3
Let the angles be x and 90-x(since we are given that the angles are complementary so their sum will be 90)
now acc to question
x-(90-x)= 14
2x= 104
x= 52
and the other angle will be 90-x= 90-52 = 38
Answer:
= + 2
Step-by-step explanation:
There is a common difference d between consecutive terms
d = 5 - 3 = 7 - 5 = 9 - 7 = 11 - 9 = 2
A recursive formula allows a term in the sequence be found by adding d to the previous term, then
= + 2 : with a₁ = 3
Answer:
3.5 by 5 by 2
That should be right.
There are two rectangles.
Multiply length times width.
-Gumina