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2 years ago

Find the inverse of the following functions f(x)=6-3x g(x)=x+2

Mathematics

2 answers:

astraxan [27]2 years ago

4 0

      f(x) = -3x + 6
    y = -3x + 6
      x = -3y + 6
   x - 6 = -3y
   -3    -3
⁻¹/₃x + 2 = y
⁻¹/₃x + 2 = f⁻¹(x)

g(x) = x + 2
   y = x + 2
   x = y + 2
x - 2 = y
x - 2 = g⁻¹(x)

Volgvan2 years ago

4 0

Problem 1:

$y=f\left( x \right) ,\\ \\ y=6-3x\\ \\ 3x=6-y\\ \\ x=\frac { 6 }{ 3 } -\frac { y }{ 3 } \\ \\ x=2-\frac { 1 }{ 3 } y\\ \\ \therefore \quad { f }^{ -1 }\left( x \right) =2-\frac { 1 }{ 3 } x$

Problem 2:

$y=g\left( x \right) ,\\ \\ y=x+2\\ \\ x=y-2\\ \\ \therefore \quad { g }^{ -1 }\left( x \right) =x-2$

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Two cylinders are similar. The radius of cylinder A is 5.6 inches. The radius of cylinder B is 1.4 inches. If the height of cyli

natita [175]

The answer is 16 inches.

Let r be a radius and h be a height of a cylinder
Since the cylinders are similar, then the ratio between their radiuses is equal to the ratio of their heights:
r1 : r2 = h1 : h2

Cylinder A:
r1 = 5.6 in
h1 = ?

Cylinder B:
r2 = 1.4 in
h2 = 4 in

5.6 : 1.4 = h1 : 4
h1 = 4 * 5.6 : 1.4
h1 = 16 inches

7 0

3 years ago

Read 2 more answers

URGENT NEED HELP< WILL GIVE BRAINLIEST!!

leonid [27]

y = 7 + 5/(x-6)

That should work.

Hope this helps!

7 0

3 years ago

(If this question is easy, I am just horrible with percentages, lol :p)

Sati [7]

55% of 20 is 11. the answer is 20

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2 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

2 years ago

CCSS Math 7

Andre45 [30]

Answer:

a = 112

b = 68

c = 68

a = 127

a=35

b=40

c=35

d=70

a= 30

b=70

c = 30

d=70

e = 130

I'll help you with the rest later

Step-by-step explanation:

a = 112 because of allied angles rule

b and c = 68 because of angles at a point

360-112-112 ÷2

2. a = 127 because of angles on a straight line rule.

180-38-15

3. d= 70, vertically opposite angle

using angles on a straight line, 180 - 70 - 40 ÷ 2

we now have the two angles and because they are vertically opposite a and c = 35

b = 40 because of vertically opposite angles

4. a=30 because 90-70

since a=30, take 90 - 30 to get b, 70

d= 70, vertically opposite angles

e = 130 because a+b+c, vertically opposite angles

7 0

2 years ago