Answer:
The answer to your question is:
x₁ = 0
x₂ = -3
x₃ = 3
Step-by-step explanation:
Equation 4x³ = 36 x
Process
4x³ - 36x = 0
4x (x² - 9) = 0
4x (x + 3)(x - 3) = 0
Finally
4x = 0 x + 3 = 0 x -3 = 0
x = 0/4 x₂ = -3 x₃ = 3
x₁ = 0
Can you please insert a diagram next time or a picture since We don't know any other information. It's completely impossible to answer this without more information.
Answer:
3/5
Step-by-step explanation:
5x= 3
x= 3/5
hope you understand the answer
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<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
Answer:
y=3/5*x, A(0,0), B(1,3/5), C(2,6/5)
Step-by-step explanation:
A(5,3)=(x1,y1)....x1=5, y1=3
m=3/5
y-y1=m(x-x1)
y-3=3/5(x-5)
y-3=3/5*x-3/5*5
y-3=3/5*x - 3
y=3/5*x-3+3
y=3/5*x
x=0, y=3/5*x, y=3/5*0=0 A(0,0)
x=1, y=3/5*1=3/5 B(1,3/5)
x=2, y=3/5*2=6/5 C(2,6/5)