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tino4ka555 [31]
3 years ago
5

I’ll CashApp someone 5$ if they can kindly answer this, I’ve been stuck on this and it’s due tonight, I will even mark you Brain

liest and give extra points please and thank you!

Mathematics
1 answer:
lisov135 [29]3 years ago
4 0

Answer: Part A is 2 and 6 Part B is 2

Step-by-step explanation:

Part A: Here is the explanation. So, you started at with the expression 3x^2+8x+4 and when you're are factoring, you have 3x^2+px+pq+4. You can substitute the p and q for 6 and 2. What they did is they replaced 8x with px+qx. To get 8x, p needs to be 6 and q needs to be 2, or the other way around. TIP: The numbers just have to add up to 8 on this one. It doesn't have to be 6 and 2.

Part B: Here is what I got so far... 3x(x+r) is 3x^2+3xr. Also, s(x+r) is sx+sr. The equation becomes, 3x^2+3xr+sx+sr. R can be 2 and s can be 2. Here is my reasoning: The original expression was 3x^2+8x+4. We already have the 3x^2, so now we need to find what the others are by determining what r and s equal. R and s can both be 2 to make four. 2x2 is 4. Let's see if it can make 8.  3xr becomes 6x and sx becomes 2x. 6x+2x is 8x.

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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?
beks73 [17]
So, how much acid is there in 6 gallons?  well is 20% acid or (20/100), so the amount of acid in it just (20/100) * 6 or 1.2, the rest is say water.

now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it?  well (90/100) * y, or 0.9y.

now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it?  (100/100) * x, or 1x or just x.

we know whatever "x" and "y" amounts are, they -> x + 6 = y

and we also know that x + 1.2 = 0.9y

\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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