1st. x times 20+100
2nd. 100+20x
3rd. 100+x times20
4th. 20x+100
5th. x20+100
Oh. It's rounding. So, for example, depending what you were told to round to, the second problem would be 760-330=430
Let <em>q</em> be the number of quarts of pure antifreeze that needs to be added to get the desired solution.
8 quarts of 40% solution contains 0.40 × 8 = 3.2 quarts of antifreeze.
The new solution would have a total volume of 8 + <em>q</em> quarts, and it would contain a total amount of 3.2 + <em>q</em> quarts of antifreeze. You want to end up with a concentration of 60% antifreeze, which means
(3.2 + <em>q</em>) / (8 + <em>q</em>) = 0.60
Solve for <em>q</em> :
3.2 + <em>q</em> = 0.60 (8 + <em>q</em>)
3.2 + <em>q</em> = 4.8 + 0.6<em>q</em>
0.4<em>q</em> = 1.6
<em>q</em> = 4
Answer: x>- 7/3
Step-by-step explanation:
Solve for x by simplifying bot sides of the equation, then isolating the variable.
For the first equation, the answer is C) completing the square.
For the second equation, the answer is B) zero product property.
For the first equation, we can easily complete the square by finding half of b and squaring it; then we can take the square root of both sides and solve the equation.
For the second equation, since it is already factored, we use the zero product property to solve it.
