Hello,
1 liter of water in normal conditions has a volume of 1dm^3=(1dm)^3=(10 cm)^3=10^3 cm^3=1000 cm^3
Answer:
61.25%
Step-by-step explanation:
How many ml of 20 percent acid should be added to pure acid
to make 70 ml of 30 percent acid?
:
let x = amt of 20% mixture
then
(70-x) = amt of pure acid
:
0.20x + (70-x) =0 .30(70)
0.20x + 70 - x = 21
= 21 - 70
-0.8x =-49
x = 61.25%
x = +61.25 ml of 20% stuff
Answer:
the equation should be corrected to fit the data of the problem. With the corrected equation a mass of 0.5 grams remains after 150 years
Step-by-step explanation:
for the mass y( in grams)
y=23* (1/2)^(t/45), t ≥ 0.
the initial mass is at t=0 , then
y= 23 grams → should be 16 grams
half-life from the equation = 45 years → should be 30 years
the correct equation should be
y=16*(1/2)^(t/30), t ≥ 0
then after 150 years → t= 150
y=16*(1/2)^(150/30)= 16*(1/2)^5 = 16/32 = 0.5 grams
then a mass of 0.5 grams remains after 150 years
Selling price of the chair = 1440 rupees
Percentage of loss made by the shopkeeper = 10%
Let us assume the price at which the shopkeeper bought the chair = x
Then
90% * x = 1440 rupees
(90/100) * x = 1440 rupees
9x/10 = 1440 rupees
9x = 1440 * 10 rupees
9x = 14400 rupees
x = 14400/9 rupees
= 1600 rupees
So the price at which the shopkeeper bought the chair was 1600 rupees. I hope the procedure is clear enough for you to inderstand.
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