The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 4.7 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?

0.94 liters antifreeze is  drained out and replaced with water to get a concentration to 50 per cent of the solution.

Step-by-step explanation:

Total amount of the solution is 4.7 liters.

It contains 70 percent antifreeze = 70% * of 4.7 = 3.29 liters of antifreeze

It contains 30 percent water = 30% * of 4.7= 1.41 liters water

For the solution to have 50 % antifreeze it must contain

4.7 * 50%= 2.35 antifreeze.

We have 3.29 liters and we want to have 2.35 liters of antifreeze

Therefore we will (3.29-2.35 =)0.94 liters drained out and replace it with water to get a concentration to 50 per cent.

12

step-by-step explanation:

a triangle's   full angle is 180 degrees so 180 - 156 = 24. 24/2=12

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.