solid line. we shade above the line
the dotted line. shadow above the line
Answer in the attachment (the first graph).
Help me please anybody. And I am in my school iPad
1 answer:
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<em>Find the equation of the line passing through (4, 5) and parallel to </em>
<em> 3x-2y=4.</em>
<em />
<em> </em>
First of all, We need to convert this equation from standard form (ax+by=c) into slope-intercept form (y=mx+b).
First, add 2y on both sides:-
Now, subtract 3x on both sides:-
Divide by 2 on both sides
Now, let's find the equation of the line that's parallel to the given line.
Remember, if lines are parallel to each other, then their slope's the same.
So the slope of the line that's parallel to the line whose equation we just found, is
Now let's write the equation in point-slope form:-
Replace numbers with letters, as follows:-
The equation above is the equation in point-slope form.
Incase you need it converted to slope intercept form, refer to the steps below ~
Multiply -3/2 times x and -4:-
Now, Add 5 on both sides:-
#TogetherWeGoFar
Answer:
Step-by-step explanation:
given a point the equation of a line with slope m that passes through the given point is
or equivalently
.
Recall that a line of the form , the y intercept is b and the x intercept is
.
So, in our case, the y intercept is and the x intercept is
.
In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph . Which means that
The slope of the tangent line is given by the derivative of the function evaluated at . Using the properties of derivatives, we get
. So evaluated at
we get
Replacing the values in our previous findings we get that the y intercept is
The x intercept is
The triangle in consideration has height and base
. So the area is
So regardless of the point we take on the graph, the area of the triangle is always 2.
FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE.
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2