A rectangular field has an area of 1,764 m2. The width of the field is 13 m more than the length. What is the perimeter of the field?
Answers
Answer:
170.
Step-by-step explanation:
We have that A = W*L where A= Area, W= width, L=length and that W = 13+L. So,
A = (13+L)*L
1764 = 13L + L^2
L^2 + 13L - 1764 = 0.
We use the general formula in the image where a=1, b=13 and c= -1764.
, this result will be negative and the problem is about length so it doesn't apply for this problem. We use L1= 36.
Then, L=36, W = 13 + 36 = 49. Therefore, the perimeter is 2L+2W = 72+98=170.
Here in France we call length the greater side of a rectangle.
So:
Let a the greater side
b the shortest
a=b+13
a*b=1764
==>b*(b+13)=1764
==>b²+13b-1764=0
==>b=(-13-85)/2 to exclude
or (b=(-13+85)/2=36 and a=36+13=49)
P=2(a+b)=2(36+49)=170 (m)
Proof:
49-36=13
36*49=1764
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Answers
Answer:
the answer is /36
Step-by-step explanation:
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b)c(x) = 2.00 + 0.50x
c) c(x) = 2.50x
d)c(x) = (2.00 + 0.5
Answers
Remember that since 2.00 is the flat fee, so there is no variable. After that, though, there is the fee of .50 which is charged every day you have a movie.
say you had the movie rented out for five days
c(5)=2.00+.50(5)
=2.00+2.5
c(5)=4.5
Hope this helped!
Answer:
The cost function that represents this scenario is c(x) = 2 + 0.50x .
Option (b) is correct .
Step-by-step explanation:
As given
Laura rents a movie for a flat fee of $2.00 plus an additional $0.50 for each night she keeps the movie.
if x equals the number of nights Laura has the movie.
Than the cost function that represents this scenario .
c(x) = Flat fee + Cost for x equals the number of nights Laura has the movie.
c(x) = 2 + x × 0.50
c(x) = 2 + 0.50x
Therefore the cost function that represents this scenario is c (x) = 2 + 0.50x .
Option (b) is correct .
Answers
solve for h
V=lwh
divide both sides by lw
V/(lw)=h
tada
Answers
x/6=tan 60°==>x/6=√3==>x=6√3
Answer B
A.
all real numbers
B.
all positive numbers
C.
all negative numbers
D.
all real numbers except 0
Answers
Answer A
There are no value to exclude with a square.