440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze
<em><u>Solution:</u></em>
Let "x" be the gallons of 80 % antifreeze added
Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze
Final mixture is x + 80
Therefore, we can frame a equation as:
"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze
Thus, we get,
x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %
Thus 440 gallons of 80 % antifreeze solution must be mixed
Answer:
the volume of the sphere is
Step-by-step explanation:
This problem bothers on the mensuration of solid shapes, sphere and cube.
Given data
Volume of cube v = 64 cubic inches
since we are dealing with a cube the height and the radius of the sphere is same as the sides of the cube,
we know that volume of cube is expressed as
![l= \sqrt[3]{64}](https://tex.z-dn.net/?f=l%3D%20%5Csqrt%5B3%5D%7B64%7D)
also diameter d=length l
Diameter d= 
Radius r = 
= 
= 
Height h=
we know that the volume of a sphere is given by
substituting into the formula we have
Answer:
p = 100(J/K - 1)
Step-by-step explanation:
When we want to make a variable the subject of an equation, we want to get that variable by itself on one side of the equation - to unwrap it from all the operations attached to it.
Here's what the process looks like for p:
X^2-10x+25 = (x-5)^2 or (x-5)(x-5)
