Answer:
-50x^2+30x-4
Step-by-step explanation:
Trace this composite figure. 2 cm 2 cm 4 cm h 5 cm 3 cm 6 cm
1 answer:
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B)
d)
9) I'm not really sure how to answer it, but I guess like 2.1, 2.2, 2.5, etc.,
10)
so {5, 6, 7, 8} is the answer
11) for some reason it won't let me insert a picture but put place
above 2, and then place
above 3, and then place
in between 2 and 3, but place it a little closer to three since
12) place
between 2 and three, but closer to two since
and also
and for number 13) the square root of 144 is 12, and the square root of 169 is 13, so any numbers between 12 and 13 will work.
I hope this helped and if not, message me and ill try to explain!
d)
9) I'm not really sure how to answer it, but I guess like 2.1, 2.2, 2.5, etc.,
10)
so {5, 6, 7, 8} is the answer
11) for some reason it won't let me insert a picture but put place
above 2, and then place
above 3, and then place
in between 2 and 3, but place it a little closer to three since
12) place
between 2 and three, but closer to two since
and also
and for number 13) the square root of 144 is 12, and the square root of 169 is 13, so any numbers between 12 and 13 will work.
I hope this helped and if not, message me and ill try to explain!
<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.
_____
<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