If 9p = 3a + 1 and 7p = 2a - 3, then which of the following expresses p in terms of a

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

4 years ago

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

Help asap pls and thx

vazorg [7]

Step-by-step explanation:

from what I can see there are 2 options to make the 2 triangles congruent :

either the opposite sides must be equal or the neighboring sides must be equal.

so, either

5x = 3x + 10

and

5x - 2 = 4x + 3

or

5x = 4x + 3

and

5x - 2 = 3x + 10

for the first

5x = 3x + 10

2x = 10

x = 5

control

5x - 2 = 4w + 3

5×5 - 2 = 4×5 + 3

23 = 23

correct.

for the second

5x = 4x + 3

x = 3

control

5×3 - 2 = 3×3 + 10

15 - 2 = 9 + 10

13 = 19

no, not a valid solution.

so, x = 5

5 0

2 years ago

Which system of linear equations has no solution?

Nesterboy [21]

Your answer is c i pretty insurance

5 0

3 years ago

For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the i

scoundrel [369]

Answer:

a) $H_0 :\mu = 4\\ H_1 : \mu \neq 4$ , n = 15 , X=3.4 , S=1.5 , α = .05

Formula : $t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}$

$t =-1.549$

p- value = 0.607(using calculator)

α = .05

p- value > α

So, we failed to reject null hypothesis

b) $H_0 :\mu = 21\\ H_1 : \mu < 21$ , n =75 , X=20.12 , S=2.1 , α = .10

Formula : $t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}$

$t =-3.6290$

p- value = 0.000412(using calculator)

α = .1

p- value< α

So, we reject null hypothesis

(c) $H_0 :\mu = 10\\ H_1 : \mu \neq 10$ , n = 36, p-value = 0.061.

Assume α = .05

p-value = 0.061.

p- value > α

So, we failed to reject null hypothesis

7 0

3 years ago