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vivado [14]
3 years ago
5

Find the unit vectors that are parallel to the tangent line to the parabola y = x2 at the point (2, 4).

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
6 0

Answer:

The unit vectors are:

\vec u_1 = \left\\\vec u_2 = \left

Step-by-step explanation:

Unit vectors that are a parallel to the tangent line have the same slope than the tangent line, thus we can find the slope of the tangent line, find directional vectors and then their corresponding unit vectors.

Finding the slope of the tangent line.

We can find the first derivative of the curve, which represents the slope of the tangent line of the curve at that point.

y = x^2 \\ y'=2x

At x = 2 we have

y'= 2(2)\\y'= 4\\

Thus we have slope m = 4, then the parallel unit vector has slope m = 4/1

Finding unit vectors.

From the slope m = 4/1, we can write it as a direction vector with x =1 and y = 4. Notice also that x = -1 and y = -4 would have given as as well slope m = 4 too.

\vec u =

Then in order to find the unit vector on that direction we can use the formula

\hat u =\cfrac{\vec u}{|\vec u|}

So finding the magnitude we get

|\vec u |= \sqrt{(1)^2+(4)^2}\\|\vec u| = \sqrt{17}

Then one unit vector that is parallel to the tangent line is

\vec u_1 = \left

And the second unit vector is

\vec u_2 = \left

The negative sign on both x and y components just tell us that we are aiming on the opposite direction as the first unit vector, yet both have the same value of the slope.

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If sample data come from a population that is not normally distributed, which
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Answer:

For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30

Step-by-step explanation:

For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30 i.e n ≥ 30 this is because the shape of a sample distribution depends on the sample size. But for  normal distribution population, the sample means would be approximately a normal distribution even if the sample size is less than 30;

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3 years ago
Which expression is equal to 15x - 6
Keith_Richards [23]
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explore how the values of algebraic expressions like100-x,5/x+5,and
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3 years ago
In tossing four fair dice, what is the probability of tossing, at most, one 3
11Alexandr11 [23.1K]

<u>Answer- </u>

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

<u>Solution-</u>

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48                    ( ∵ xxxx )

P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38                    ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0
3 years ago
Find the surface area and volume.
Rufina [12.5K]

Answer:

surface area=2816 m² & volume =205.33m³

Step-by-step explanation:

here

diameter of base (d) -28 m

radius(r) = d/2=14 m

vertical height of cone (h)-48 m

slant height of cone (l)-50 m

surface area (TSA)-?

volume (v)-?

we know

TSA=(pie*r*l)+pie*r*r

=pie*r(l+r)

=22/7* 14(50+14)

=22*2(64)

=22*128

=2816 m²

again

v=1/3 pie*r*r

=1/3*22/7*14*14

=1/3*22*2*14

=205.33 m³

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2 years ago
Tell whether the lines for the pair of equations are parallel, perpendicular, or neither.
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