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Fudgin [204]
3 years ago
15

Create a graph of y=2x−6. Construct a graph corresponding to the linear equation y=2x−6.

Chemistry
1 answer:
yan [13]3 years ago
7 0
The graph is a straight line with this characteristics:

slope: the coefficient that accompanies the x variable, so it is 2.

x-axis intercept: find from y = 0 => 2x - 6 = 0 => x = 3

y - axis intercept: find from x = 0 => y = 2(0) - 6 = 6.

With that information you can draw you graph.

Just in case you need more information, you can build a table with several values of x and y:

x        y = 2x - 6

-5      2(-5) - 6 = -16

-4      2(-4) - 6 = -14

-3      2(-3) - 6 = -12

-2      2(-2) - 6 = -10

-1      2(-1) - 6 = -8

0        2(0) - 6 = -6

1        2(1) - 6 = -4

2        -2
 
3         0

4         2

5         4

Now you have plenty information to draw the graph.

6         6
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A particle has a mass of 2 × 10-30 kg and an uncertainty in its velocity of 102 m/s. What is the minimum possible uncertainty in
notka56 [123]

m = mass of the particle = 2 x 10⁻³⁰ kg

Δv = uncertainty in velocity of the particle = 10² m/s

Δx = uncertainty in position of the particle = ?

h = plank's constant = 6.63 x 10⁻³⁴ J-s

uncertainty principle is given as

m Δx Δv ≥ h/(4π)

inserting the values

(2 x 10⁻³⁰) Δx (10²) ≥ (6.63 x 10⁻³⁴)/(4(3.14))

Δx ≥ 2.64 x 10⁻⁷ m

so minimum possible uncertainty is 2.64 x 10⁻⁷ m

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3 years ago
Nic bought 12 pens for 2$ each and 12pads of paper for 3$ each. Write two expressions for how nic could calculate his total.Writ
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12 pens for writing chemical formulas cost
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12 pads cost
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1)
12*2=24
12*3=36
2)
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A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.
inna [77]
PKa= 4.9 therefore ka= 10^-4.9=  1.259x10^-5

ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}

[CH3CH2COO^-]
= 0.05
[CH3CH2COOH]= 0.10

Therefore 1.259x10^-5 = \frac{[H^+][0.05]}{[0.1]}
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore [H^+] =  \frac{(1.259*10^-5)(0.1)}{0.05}

Therefore [H^+]= 2.513*10^-5

pH= -log [H^+] = -log(2.513*10^-5)= 4.59.
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