SOLUTION: i actually have a few questions some aren't really about the topic...a few being 5... 1. Two kilograms of iron was melted with 7 kilograms of other metals to make the alloy. If

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Question 72980This question is from textbook Algebra 2
: i actually have a few questions some aren't really about the topic...a few being 5...
1. Two kilograms of iron was melted with 7 kilograms of other metals to make the alloy. If 1440 kilograms of the alloy was required, how many kilograms of iron should be used?
2. Charles and Matthew knew that the formula for sulfuric acid was H2SO. If they haad 196 grams of sulfuric acid, what was the weight of the sulfur(H,1; S,32; O,16)?
3. The ratio of the two numbers was 7 to 2. When Sir Richard and MArion multiplied the denominator by 10, they found that the result was 84 greater than twice the numerator. What were the numbers?
4. Solve: 5x-2 over(/) 3 -x over(/) 4 = 7
5. 1 over (/) x+3 + 3x over(/) x+2 + 2x+1 over(/) x squared + 5x + 6 This question is from textbook Algebra 2

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1. Two kilograms of iron was melted with 7 kilograms of other metals to make the alloy. If 1440 kilograms of the alloy was required, how many kilograms of iron should be used?
:
the total is 2 + 7 = 9: Relationship of iron to the total: 2:9
Let I = kilos of iron
:
2%2F9 =I%2F1440
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Cross multiply:
9I = 2(1440)
9I = 2880
I = 2880/9
I = 320 kilos of iron in 1440 kilos of alloy
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2. Charles and Matthew knew that the formula for sulfuric acid was H2SO. If they had 196 grams of sulfuric acid, what was the weight of the sulfur(H,1; S,32; O,16)?
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The relationship of a sulfur weight to the total: 32:(2+32+16) = 32:50
:
32%2F50 = S%2F196
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Cross multiply:
50S = 6272
s = 6272/50
s = 125.44 grams of sulfur in 196 grams of H2SO
:
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3. The ratio of the two numbers was 7 to 2. When Sir Richard and MArion multiplied the denominator by 10, they found that the result was 84 greater than twice the numerator. What were the numbers?
:
"The ratio of the two numbers was 7 to 2"
Let x = the multiplier
:
%287x%29%2F%282x%29
:
"Multiply the denominator (2x) by 10.
Result is 84 greater than twice the numerator (7x)"
10(2x) = 2(7x) + 84
20x = 14x + 84
20x - 14x = 84
6x = 84
x = 14 is the multiplier
:
Therefore the numbers are: (14*7) and 14(2) or 98 and 28
:
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4. Solve: 5x-2 over(/) 3 -x over(/) 4 = 7
:
%28%285x-2%29%29%2F3 - x%2F4 = 7
:
Multiplying equation by 12 gets rid of the denominators:
4(5x-2) - 3(x) = 12(7)
20x - 8 - 3x = 84
20x - 3x = 84 + 8
17x = 92
x = 92/17
:
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5. 1 over (/) x+3 + 3x over(/) x+2 + 2x+1 over(/) x squared + 5x + 6
:
The way I understand it:
1%2F%28x%2B3%29+%2B+3x%2F%28x%2B2%29+%2B+%282x%2B1%29%2F%28x%5E2+%2B+5x+%2B+6%29
The last denominator can be factored to the same values as the other 2 denominators:
1%2F%28x%2B3%29+%2B+3x%2F%28x%2B2%29+%2B+%282x%2B1%29%2F%28%28x%2B2%29%28x%2B3%29%29
:
(x+3)(x+2) is the common denominator
%281%28x%2B2%29+%2B+3x%28x%2B3%29+%2B+%282x%2B1%29%29%2F%28%28x%2B2%29%28x%2B3%29%29 = %28x+%2B+2+%2B+3x%5E2+%2B+9x+%2B+2x+%2B+1+%29%2F%28%28x%2B3%29%28x%2B2%29%29 = %28+3x%5E2+%2B+12x+%2B+3+%29%2F%28%28x%2B3%29%28x%2B2%29%29 = %28+3%28x%5E2+%2B+4x+%2B+1+%29%29%2F%28%28x%2B3%29%28x%2B2%29%29
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That's about as far as you can go with it