Please show the work. 5256÷52, 5624÷46, 756÷36, 2760÷12. It's a lot to do, so I'm making this 40 points.
1 answer:
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Answer:
Step-by-step explanation:
we need to find the equation of the parabola with vertex (-5,3), with axis parallel to the x-axis.
general equation of horizontal parabola is
where (h,k) is the vertex
(-5,3) is the vertex. h=-5 and k=3. plug in the values in the general equation
Now find out the value of 'a' using x intercept (4,0)
Add 5 on both sides
a=1
the equation becomes
<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