A short-cut to accurately evaluate the given expression above is using a scientific calculator where one can include integrals and evaluate using limits. In this case, using a calculator, the answer is equal to 0.2679. One can verify this by integrating truly letting 1-x^2 as u and use its du to be substituted in the numerator
13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed
<u>Step-by-step explanation:</u>
Total mass of acid required= 85 ounces
Let the mass of 72% acid be 'a'
Let the mass of 25% acid be 'b'
a + b = 85
b = 85-a
85(40/100) = a(72/100) + b(25/100)
85(2/5) = a(72/100) +(85- a) (25/100)
34 = (72a/100) + (2125/100) - (25a/100)
34 - (2125/100) = (72a + 25a) /100
(3400-2125)/100 = 97a /100
97a = 1275
a = 13.15 ounces
b = 85 - 13.14
b = 71.85 ounces
13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed
Answer:
12y-24
thats what the answer is to your problem
Answer:
Calcium carbonate reacts w/stomach acid according to the following chemical equation.
CaCO3+2HCl(aq)-> CaCl2(aq)+H2O(l)+CO2(g)
Answer:
Step-by-step explanation:
Multiply both sides by 3/4.
(3/4)(4/3)x = (3/4)8
x = 6
