Say I have a graph. The points plotted make a straight line, but the textbook did not draw a straight line, they drew dots. Woul
1 answer:
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Answer:
The correct option is (1,2), (0, 1), (1, 2), (2, 4).
Step-by-step explanation:
Consider the provided ordered pairs.
The exponential function is in the form of
Where, a is the initial amount and b is the rate of change b>0,
The graph of each exponential function passes through the point (0, a),
and
Therefore, Option 1 and 2 and 4 are false.
Hence, the correct option is (1,2), (0, 1), (1, 2), (2, 4).
Answer:
left
Step-by-step explanation:
The question says she used of it, so she probably used
of it. I'm assuming this is an error in the question, and it should be asking how much is left. To solve for that:
-
make denominators equal
( ×
) - (
×
)
-
a. The equation of the tangent at (1,3) is y = -x/3 + 10/3
b. The equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
<h3>a. How to find the equation of the tangent at (1, 3)?</h3>Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.
So, x² + y² = 10
d(x² + y²)/dx = d10/dx
dx²/dx + dy²/dx = 0
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y
At (1,3), dy/dx = -1/3
Using the equation of a straight line in slope-point form, we have
m = (y - y₁)/(x - x₁) where
- m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
- (x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
-1/3 = (y - 3)/(x - 1)
-(x - 1) = 3(y - 3)
-x + 1 = 3y - 9
3y = -x + 1 + 9
3y = -x + 10
3y + x = 10
y = -x/3 + 10/3
So, the equation of the tangent at (1,3) is y = -x/3 + 10/3
<h3>b. The equation of the normal at the point (1, 3)</h3>Since the tangent and normal line are perpendicular at the point, for two perpendicular line,
mm' = -1 where
- m = gradient of tangent = -1/3 and
- m' = gradient of normal
So, m' = -1/m
= -1/(-1/3)
= 3
Using the equation of a straight line in slope-point form, we have
m' = (y - y₁)/(x - x₁) where
- m' = gradient of normal at (1, 3) and (
- x₁, y₁) = (1,3)
So, m = (y - y₁)/(x - x₁)
3 = (y - 3)/(x - 1)
3(x - 1) = (y - 3)
3x - 3 = y - 3
y = 3x - 3 + 3
y = 3x + 0
y = 3x
So, the equation of the normal at (1,3) is y = 3x
c. Find the graph in the attachment
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