Answer:
You need to show a graph
Step-by-step explanation:
Answer:
Length = 2x+y cm and since it's a rectangle,
2x+y=3x-y ---------------- (i)
width = 2x-3 cm
It's perimeter,
2(2x+y+2x-3)=120 ---------------- (ii)
Solving both equations,
x = 14 cm
y = 7 cm
so length is, 2×14+7 = 35 cm
and width is, 2×14-3 = 25 cm
so area will be, 35×25 = 875 cm²
Answered by GAUTHMATH
We want to know when h is equal to 0. This represents when the height is 0 and thus, hits the ground.
Then, 0 = 70t - 5t²
0 = 5t² - 70t
0 = 5t(t - 14)
t = 0 and 14.
So, at t = 0, the initial height is 0 and it hits the ground again after 14 seconds.
Answer:
Step-by-step explanation:
Given that the owner of a motel has 2900 m of fencing and wants to enclose a rectangular plot of land that borders a straight highway.
Fencing is used for 2times length and 1 width if highway side is taken as width
So we have 2l+w = 2900
Or w = 2900-2l
Area of the rectangular region = lw
![A(l) = l(2900-2l) = 2900l-2l^2\\](https://tex.z-dn.net/?f=A%28l%29%20%3D%20l%282900-2l%29%20%3D%202900l-2l%5E2%5C%5C)
Use derivative test to find the maximum
![A'(l) = 2900-4l\\A"(l) = -4](https://tex.z-dn.net/?f=A%27%28l%29%20%3D%202900-4l%5C%5CA%22%28l%29%20%3D%20-4%3C0)
So maximum when I derivative =0
i.e when ![l =\frac{2900}{4} =725](https://tex.z-dn.net/?f=l%20%3D%5Cfrac%7B2900%7D%7B4%7D%20%3D725)
Largest area = A(725)
= ![725(2900-2*725)\\= 1051250](https://tex.z-dn.net/?f=725%282900-2%2A725%29%5C%5C%3D%201051250)
1051250 sqm is area maximum
Answer:
True
Step-by-step explanation: