Write an equation in slope-intercept form with the given slope and y-intercept. m = 1 , b = -1
1 answer:
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Problem 13a
Each column of the table represents an (x,y) pairing. The first column has x = 0 and y = 2 so (0,2) is one point on this line. Since x = 0 here, this means that we have the y intercept. The y intercept is 2. We'll use this later.
Another point on this line is (3,8) which is drawn from the second column.
Use the two points (0,2) and (3,8) to find the slope m
m = (y2-y1)/(x2-x1)
m = (8-2)/(3-0)
m = 6/3
m = 2
Since the slope is 2 and the y intercept is 2 (see above), this means m = 2 and b = 2
Plug those into y = mx+b and we get y = 2x+2
Answer: y = 2x+2
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Problem 13b
(0,20) is on the line as the first column shows. The y intercept is 20 since x = 0 here. So b = 20
Another point on this line is (3,8) as drawn from the second column
Slope:
m = (y2-y1)/(x2-x1)
m = (8-20)/(3-0)
m = -12/3
m = -4
Use the slope (m = -4) and the y intercept (b = 20) to get...
y = mx+b
y = -4x+b ... replace m with -4
y = -4x+20 ... replace b with 20
Answer: y = -4x+20
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Problem 13c
(2,5) and (4,8) are on the line. These two points are drawn from columns 1 and 2 respectively.
Slope
m = (y2-y1)/(x2-x1)
m = (8-5)/(4-2)
m = 3/2
m = 1.5
Note: the fraction 3/2 is equivalent to the decimal form 1.5
Use the slope m = 1.5 along with (x,y) = (2,5) to find the value of b
y = mx+b
y = 1.5x+b ... m is replaced with 1.5
5 = 1.5*2+b ... plug in (x,y) = (2,5)
5 = 3+b
5-3 = 3+b-3
2 = b
b = 2
So we can now update y = mx+b to y = 1.5x+2
Answer: y = 1.5x+2
Note: If your teacher wants a fraction instead of a decimal (for the slope), then you would write
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Problem 13d
(0,20) is on the line. So the y intercept is 20, indicating that b = 20
(3,11) is also on the line
Again as done with 13a-13c, I'm only looking at the first two columns.
Slope
m = (y2-y1)/(x2-x1)
m = (11-20)/(3-0)
m = -9/3
m = -3
Therefore
y = mx+b
turns int
y = -3x+20
after replacing m and b with the values we found earlier
Answer: y = -3x+20
Each column of the table represents an (x,y) pairing. The first column has x = 0 and y = 2 so (0,2) is one point on this line. Since x = 0 here, this means that we have the y intercept. The y intercept is 2. We'll use this later.
Another point on this line is (3,8) which is drawn from the second column.
Use the two points (0,2) and (3,8) to find the slope m
m = (y2-y1)/(x2-x1)
m = (8-2)/(3-0)
m = 6/3
m = 2
Since the slope is 2 and the y intercept is 2 (see above), this means m = 2 and b = 2
Plug those into y = mx+b and we get y = 2x+2
Answer: y = 2x+2
==============================================
Problem 13b
(0,20) is on the line as the first column shows. The y intercept is 20 since x = 0 here. So b = 20
Another point on this line is (3,8) as drawn from the second column
Slope:
m = (y2-y1)/(x2-x1)
m = (8-20)/(3-0)
m = -12/3
m = -4
Use the slope (m = -4) and the y intercept (b = 20) to get...
y = mx+b
y = -4x+b ... replace m with -4
y = -4x+20 ... replace b with 20
Answer: y = -4x+20
==============================================
Problem 13c
(2,5) and (4,8) are on the line. These two points are drawn from columns 1 and 2 respectively.
Slope
m = (y2-y1)/(x2-x1)
m = (8-5)/(4-2)
m = 3/2
m = 1.5
Note: the fraction 3/2 is equivalent to the decimal form 1.5
Use the slope m = 1.5 along with (x,y) = (2,5) to find the value of b
y = mx+b
y = 1.5x+b ... m is replaced with 1.5
5 = 1.5*2+b ... plug in (x,y) = (2,5)
5 = 3+b
5-3 = 3+b-3
2 = b
b = 2
So we can now update y = mx+b to y = 1.5x+2
Answer: y = 1.5x+2
Note: If your teacher wants a fraction instead of a decimal (for the slope), then you would write
==============================================
Problem 13d
(0,20) is on the line. So the y intercept is 20, indicating that b = 20
(3,11) is also on the line
Again as done with 13a-13c, I'm only looking at the first two columns.
Slope
m = (y2-y1)/(x2-x1)
m = (11-20)/(3-0)
m = -9/3
m = -3
Therefore
y = mx+b
turns int
y = -3x+20
after replacing m and b with the values we found earlier
Answer: y = -3x+20
Answer: See the graph attached.
Step-by-step explanation:
1. To solve this exercise you must plot the points given in the table of the problem, which are shown below:
(-4,1), (0,2), (3,3)
2. You must plot the first value of each ordered pair on the x-axis.
3. You must plot the second value of each ordered pair on the y-axis.
4. Therefore, when you plot them, you obtain the graph shown attached.
Answer:
The draining time when only the big drain is opened is 2.303 hours.
The draining time when only the small drain is opened is 5.303 hours.
Step-by-step explanation:
From Physics, we know that volume flow rate (), measured in liters per hour, is directly proportional to draining time (
), measured in hours. That is:
(Eq. 1)
Where is the proportionality constant, measured in liters.
From statement, we have the following three expressions:
(i) <em>Large and small drains are opened</em>
(Eq. 2)
(ii) <em>Only the small drain is opened</em>
(Eq. 3)
(iii) <em>Only the big drain is opened</em>
(Eq. 4)
By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:
(Eq. 5)
Whose roots are determined by the Quadratic Formula:
and
Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:
The draining time when only the small drain is opened is 5.303 hours.