Answer:
f(x) -> 2 as x -> -∞ and f(x) -> ∞ as x -> ∞
Step-by-step explanation:
f(x) -> 2 as x -> -∞ and f(x) -> ∞ as x -> ∞
*this is because the graph shows that the y value cannot pass by 2 as the x value constantly decreases and as x is increasing, there is just an arrow showing that the graph is constantly going up and therefore going to ∞
Y=2X
Looking at the line you can see that it is moving up from left to right, making it positive. So -2X and -12X are incorrect.
Next we can find the slope by putting rise over run. Chose a point on the line, I’ll use (2,4).
On this point we rise to 4 and run over to 2.
So for rise over run we have 4/2 which is 2.
So y=2x is correct.
Answer:
B
Step-by-step explanation:
A graph skewed right would have LESS dots on the right side
Hope this helped!
<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.
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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
The bottom choice is not equal to the original expression because it has an extra x