Answer:
23.96g
Step-by-step explanation:
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Inequality:
v>83
This inequality says that
v is greater than $83.
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Answer:
Trinomial
Step-by-step explanation:
yes they're considered terms
Answer:
y = 0.5 cosine (4 (x - pi/2)) - 2
Step-by-step explanation:
Taking the general form:
y = A cosine (Bx - Cπ)) + D
In the following case. the constants are:
y = 0.5 cosine (4x - 2π)) - 2
A: 0.5
B: 4
C: 2π
D: -2
The range of this function is:
range = [-|A|+D, |A|+D]
range = [-0.5-2, 0.5-2]
range = [-2.5, -1.5]
Which coincides with "It has a maximum at negative 1.5 and a minimum at negative 2.5"
At x = 0, the function value is:
y = 0.5 cosine (4(0) - 2π)) - 2
y = 0.5 - 2 = -1.5
As indicated in "a curve crosses the y-axis at y = negative 1.5"
The period of the function is:
period: 2π/B
period = 2π/4 = π/2 or 2 cycles at π
as described in "It goes through 2 cycles at pi."
<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