0.3 then 0.13 then 0.19 then 0.31
have a nice day ;)
This is a nice "rates of change" problem from Calculus.
Let the length and width of the rect. be L and W. We are given the following info:
dL/dt = 6 cm/s; dW/dt = 4 cm/s; L = 11 cm and W = 5 cm.
The area of the rect. is A = L*W. Differentiating,
dA/dt = L(dW/dt) + W(dL/dt).
Subst. the given info: dA/dt = (11 cm)(4 cm/sec) + (5 cm)(6 cm/sec).
Just evaluate this to find dA/dt: (44 + 30) cm^2/sec = 76 cm^2/sec
Answer:
The value of y is -2
Therefore the given point (8,y) becomes (8,-2)
Step-by-step explanation:
Step-by-step explanation:
Given that the line through the points (8, y) and (9, 1) and slope m is 3
To find y to the line through the points (8, y) and (9, 1) :
The formula for slope is ![m=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
Let
,
be the given points (8, y) and (9, 1) respectively
Now substitute the values of the points and m=3 in the formula
![3=\frac{1-y}{9-8}](https://tex.z-dn.net/?f=3%3D%5Cfrac%7B1-y%7D%7B9-8%7D)
![3=\frac{1-y}{1}](https://tex.z-dn.net/?f=3%3D%5Cfrac%7B1-y%7D%7B1%7D)
![3=1-y](https://tex.z-dn.net/?f=3%3D1-y)
![3-(1-y)=1-y-(1-y)](https://tex.z-dn.net/?f=3-%281-y%29%3D1-y-%281-y%29)
![3-1+y=1-y-1+y](https://tex.z-dn.net/?f=3-1%2By%3D1-y-1%2By)
( adding the like terms )
![2+y-2=0-2](https://tex.z-dn.net/?f=2%2By-2%3D0-2)
Therefore y=-2
Therefore the value of y is -2
Therefore the given point (8,y) becomes (8,-2)
Answer:
w = 18
Step-by-step explanation:
Given
(w + 18) = 12 ← distribute parenthesis on left side
w + 6 = 12 ( subtract 6 from both sides )
w = 6 ( multiply both sides by 3 to clear the fraction )
w = 18
Step-by-step explanation:
2. Amount of soil provided.
In order to give different pH values of the soil, the amount of soil can directly affect this independent variable.