Let x gallons be the amount of pure antifreeze that should be added to the 30% solution to produce a solution that is 65% antifreeze. Then the total amount of antifreeze solution will be x+3 gallons.
There are 30% of pure antifreeze in 3 gallons of solution, then
3 gallons - 100%,
a gallons - 30%,
where a gallons is the amount of pure antifreeze in given solution.
Mathematically,

Now in new solution there will be x+0.9 gallons of pure antefreeze.
x+3 gallons - 100%,
x+0.9 - 65%
or

Answer: he should add 3 gallons of pure antifreeze.
3675. You multiply 105 700, then you multiply the answer by 0.05.
Answer:
x = -3 x=2
Step-by-step explanation:
x^2 + x - 6 = 0
Factor
What two numbers multiply to -6 and add to 1
3*-2 =-6
3+-2 =1
(x+3) (x-2) = 0
Using the zero product property
x+3 = 0 x-2 = 0
x = -3 x=2
This question has to do with making an approximation based on the diagram of the angle. Angle KLM is an obtuse angle meaning it is larger then 90° but is also less than 180°. The only choices that lie within this range are choices B and D. However, it is safe to assume that angle KLM is closer to 180° than to 90° based on the diagram, therefore the answer must be choice B.
I hope this helps.
Answer:
For
, x = 2, or x = - 2.
Step-by-step explanation:
Here, the given expression is :

Now, using the ALGEBRAIC IDENTITY:

Comparing this with the above expression, we get

⇒Either (x-2) = 0 , or ( x + 2) = 0
So, if ( x- 2) = 0 ⇒ x = 2
and if ( x + 2) = 0 ⇒ x = -2
Hence, for
, x = 2, or x = - 2.