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Jlenok [28]
3 years ago
9

what is the answer to the multiplication problem. Jhon has 20 fish if he lets 6 fish go how many fish dose Jhon have?

Mathematics
1 answer:
spayn [35]3 years ago
6 0
Isn't this a subtraction problem, not a multiplication problem?

If John starts out with 20 fish and lets 6 go, he still has (20-6), or 14, fish.
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y = -(2/3)x + 10

Step-by-step explanation:

1. The first thing to note is that when two lines are parallel, their gradients are the same. This means that the line that we are looking for will have the same gradient as that of the line 2x = 1 - 3y.

Thus, what we must first do is find the gradient of the line 2x = 1 - 3y; one way to do this is to write it in the form y = mx + c, where m is the gradient and c is the y-intercept.

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2x + 3y = 1 (Add 3y to both sides)

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3y = -2x + 1 (Swap +1 and -2x)

y = -(2/3)x + 1/3 (Divide both sides by 3)

From the equation above, we can see that the gradient is -2/3.

2. We are now aware that the gradient of the line we are looking for is -2/3 and that there is a point on the line with coordinates (9, 4).

In order to find the equation of the line, we can substitute these values into the equation y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the gradient.

In our case, m = -2/3, (x1, y1) = (9, 4). Thus, we get:

y - 4 = -2/3(x - 9)

y - 4 = -(2/3)x + 6 (Expand -2/3(x - 9) ; note that a negative value multiplied by a negative value is a positive value)

y - 4 + 4 = -(2/3)x + 6 + 4 (Add 4 to both sides)

y = -(2/3)x + 10

Thus, the equation of the line that is parallel to the line 2x = 1 - 3y, is y = -(2/3)x + 10.

Note that this can be written in many different ways, for example it may also be written as 3y = -2x + 30 (in which case, we have simply removed the fraction in the equation by multiplying both sides by 3), or 2x = 30 - 3y (which mimics the format of the equation in the question). These are all valid answers, it just depends on whether the question requires you to write it in a particular form.

Hope that helps, but if you have any further questions about the problem or answer please feel free to comment below :)

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