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Gwar [14]
3 years ago
5

Can someone help me with this I don't understand.

Mathematics
1 answer:
Troyanec [42]3 years ago
3 0
First step, write down the expression on one line, using necessary parentheses. 

(2^8*5^(-5)*19^0)^(-2) * (5^(-2)/2^3)^4 *2^28

The apply the PEMDAS rule to simplify according to priority:
1. Parentheses
2. Exponentiation
3. Multiplication and Division (left to right)
4. Addition and Subtraction (left to right)

For exponentiation, it is convenient to convert all negative exponents to positive by taking the reciprocal, for example,
a^(-2)= 1/(a^2)
So
(2^8*5^(-5)*19^0)^(-2) * (5^(-2)/2^3)^4 *2^28
=1/[(2^8*5^(-5)*19^0)^(2)] * (1/[5^2*2^3])^4 *2^28
=5^5/[(2^8*19^0)^(2)] * (1/[5^2*2^3])^4 *2^28
Remember positive numbers raised to power zero equals 1, e.g.
a^0=a for any number a.
so 19^0=1
=5^10/[(2^8)^(2)] * (1/[5^2*2^3])^4 *2^28
Now, powers raised to a power equals the product of the powers, e.g.
(2^8)^2=2^16, (5^2)^4=5^8, (2^3)^4=2^12...
continuing,
=5^10/[2^16] * 1/[5^8*2^12] *2^28
When we have exponents to the same base in the denominator and numerator, we subtract exponents, e.g.
5^5/5^8=5^(5-8)=5^(-3), 2^28/[2^16*2^12]=2^(28-16-12)=2^0=1
so we get rid of many terms this way
continuing
=5^(10-8)*2^(28-16-12)
=5^2*2^0
=25*1
=25
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How much pure acid should be mixed with 6 gallons of a 20% acid solution in order to get a 90% acid solution?
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\bf \begin{array}{lccclll}
&\stackrel{gallons}{acid}&\stackrel{acid~\%}{quantity}&\stackrel{acid~gallons}{quantity}\\
&------&------&------\\
\textit{pure acid}&x&1.00&x\\
\textit{20\% sol'n}&6&0.20&1.2\\
------&------&------&------\\\
mixture&y&0.90&0.9y
\end{array}
\\\\\\
\begin{cases}
x+6=\boxed{y}\\
x+1.2=0.9y\\
----------\\
x+1.2=0.9\left( \boxed{x+6} \right)
\end{cases}
\\\\\\
x+1.2=0.9x+5.4\implies x-0.9x=5.4-1.2\implies 0.1x=4.2
\\\\\\
x=\cfrac{4.2}{0.1}\implies x=\stackrel{gallons}{42}
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