Answer:
50-2.5t=w. no, the water will all have drained by then.
Step-by-step explanation:
2.5 per min so
min=t
therefore 2.5t
theres 50 quarts of water
50-2.5t = w (water left in the tank)
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part 2,
2.5*30 = 75, and 50<75
this is impossible.
What is y={-\dfrac{1}{3}}x-9y=− 3 1 x−9y, equals, minus, start fraction, 1, divided by, 3, end fraction, x, minus, 9 written i
Kobotan [32]
Answer:
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Step-by-step explanation:
We are given that

We have to find the standard form of given equation


By using multiplication property of equality
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We know that
Standard form of equation

Therefore, the standard form of given equation is given by
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Answer:
Step-by-step explanation:
f(x) = √(5 + 4)
f(x) = √(9) = √(3 · 3) = 3
Answer:
good byy
Step-by-step explanation:
sweet dreams.
..
<h3>
Answer:</h3>
- <u>20</u> kg of 20%
- <u>80</u> kg of 60%
<h3>
Step-by-step explanation:</h3>
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