Convert 1.004 to a percent. Enter your answer in the box. HELLPPP!!!!
1 answer:
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<h3>Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
I like to use a little X diagram to work mixture problems like this. The constituent concentrations are on the left; the desired mix is in the middle, and the right legs of the X show the differences along the diagonal. These are the ratio numbers for the constituents. Reducing the ratio 32:8 gives 4:1, which totals 5 "ratio units". We need a total of 100 kg of alloy, so each "ratio unit" stands for 100 kg/5 = 20 kg of constituent.
That is, we need 80 kg of 60% alloy and 20 kg of 20% alloy for the product.
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<em>Using an equation</em>
If you want to write an equation for the amount of contributing alloy, it works best to let a variable represent the quantity of the highest-concentration contributor, the 60% alloy. Using x for the quantity of that (in kg), the amount of copper in the final alloy is ...
... 0.60x + 0.20(100 -x) = 0.52·100
... 0.40x = 32 . . . . . . . . . . .collect terms, subtract 20
... x = 32/0.40 = 80 . . . . . kg of 60% alloy
... (100 -80) = 20 . . . . . . . .kg of 20% alloy
Ada is <u>incorrect</u>, as the length of diagonal SQ is not two times the length of diagonal OM. The theorem used is the Pythagoras Theorem.
For Rectangle PQRS:-
By Pythagoras Theorem, in right triangle PQS,
SQ² = PQ² + PS²,
or, SQ² = 8² + 16² sq. inches,
or, SQ² = 64 + 256 sq. inches,
or, SQ = √320 sq. inches,
or, SQ = 8√5 inches.
Thus, the length of diagonal SQ, of rectangle PQRS is 8√5 inches.
For Square LMNO:-
By Pythagoras Theorem, in right triangle LMO,
OM² = LM² + LO²,
or, OM² = 8² + 8² sq. inches,
or, OM² = 64 + 64 sq. inches,
or, OM = √128 sq. inches,
or, OM = 8√2 inches.
Thus, the length of diagonal OM, of square LMNO is 8√2 inches.
We know that 2*OM ≠ SQ, as 2*8√2 ≈ 8√5.
Thus, Ada is <u>incorrect</u>, as the length of diagonal SQ is not two times the length of diagonal OM. The theorem used is the Pythagoras Theorem.
Learn more about the diagonal of square and rectangle at
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