Answer How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution. ... x=15 liters 15*(.20 pure acid)=3 liters 30*(.50 pure acid)=15 liters That is 18 liters pure acid That is 45 liters solution *0.45 pure acid=18 liters.
As more pure acid is added, the concentration of acid approaches 0.
First, let me do the Mathematical part of that, and then I shall explain the theory behind it.
Mathematical part:
We are going to multiply 513 with 46. So the two partial products that we are going to choose are 40 and 6.
Multiply 513 with 6 first.
513
x46
--------------------------
18 (as 6*3 = 18)
60 (as 6*10 = 60; In 513, the digit at tenths place is 1, so 1*10=10)
3000 (as 6*500 = 3000; In 513, 5 is at hundredth place, so 5*100=500)
120 (as 40*3 = 120; since 4 is at the tenth place, so 4*10=40)
400 (as 40*10 = 400)
20000 (as 40*500 = 20000)
--------------------------
23598 (Add all of them)
Theory:
As you can see above that we have chosen the two partial products individually which are 6 and 40. Since 4 in 46 is in tenth place, we have to consider it 40 (since 4*10 = 40). One by one, we first multiply 6 with 513. Then we move to the tenth place, and multiply 513 with 40. At the end, we have added all the results we found after multiplication.
Check: If we check the multiplication result by using the calculator, we would get the same result (23598).
Another Method (instant):
513 * (40+6) = (513*40) + (513*6) = 23598.
Solve for x.
(3x+1)2-100=0
(3x+1)2=100
3x+1=100/2
3x=50-1
x=49/3
x=16.33333
Hope that helps. If you have any other questions feel free to ask me
Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
The arc length of AB is 8 m (app.)
Explanation:
Given that the radius of the circle is 8 m.
The central angle is 60°
We need to determine the arc length of AB
The arc length of AB can be determined using the formula,
Substituting central angle = 60° and circumference = 2πr in the above formula, we get,
Simplifying the terms, we get,
Dividing, we get,
Hence, the arc length is approximately equal to 8.
Therefore, the arc length of AB is 8 m
