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3 years ago

HELP NEED HELP PLEASE

Mathematics

1 answer:

inn [45]3 years ago

7 0

Answer:

the 1st and 3rd answer choices!

Step-by-step explanation:

hope this helps!!

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Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona

andrey2020 [161]

Answer:

a. v.v' = v₁v₂ - v₁v₂ = 0 b. (20, -21)/29 and (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' = (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

= v₁v₂ - v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0

3 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago

Which statement must be true about line TU?

just olya [345]

Line tu is parallel to rs

5 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the graph of the function f(x) = 2sinx at the point where x = pi/3.

BigorU [14]

The equation of the tangent line of f(x) is given by y = f'(x)x + c
f(x) = 2sin x
f'(x) = 2cos x

At x = pi/3
f'(pi/3) = 2 cos(pi/3) = 1
f(pi/3) = 2 sin(pi/3) = √3
√3 = 1(π/3) + c
c = √3 - π/3

Therefore, required equation is
y = x + √3 - π/3

3 0

3 years ago

What is the surface area of the rectangular prism below 7, 7, 14

Julli [10]

Answer:

Surface Area A = 490 cm²

Step-by-step explanation:

Surface area of a prism A = 2(w l +h l + h w)

where l = length

w = width

h = height

l = 7 cm; w = 7 cm ; h = 14 cm

Surface Area = 2 { ( 7 × 7) + ( 14 × 7) + ( 14 × 7)}

A = 2 { 49 + 98 + 98}

A = 2 × 245

A = 490 cm²

8 0

3 years ago