Write the point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 include all of you
1 answer:
<u>Answer: </u>
The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0
<u>Solution: </u>
The point slope form of the line that passes through the points and parallel to the line with slope “m” is given as
--- eqn 1
Where “m” is the slope of the line. are the points that passes through the line.
From question, given that slope “m” = -3
Given that the line passes through the points (6,1).Hence we get
By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out.
y – 1 = -3(x – 6)
y – 1 = -3x +18
On rearranging the terms, we get
3x + y -1 – 18 = 0
3x + y – 19 = 0
Hence the point slope form of given line is 3x + y – 19 = 0
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The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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Complete Question
(Image Attached)
Answer:
-2x-2
For a line equation for the form of y = mx +b, the slope is m
m= -2
The function has no undefined points nor domain
constraints. There fore, the domain is
domain: -∞<x<∞
The range of polynomial with odd degrees is all the real numbers
range: -∞<f(x)<∞
Step-by-step explanation:
graph is inserted
Answer:
The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
93% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).