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3 years ago

Find the constant variation for this equation. R=.51/d^2

1 answer:

7 0

A constant of variation is part of a direct relation - linear function. The equation given is an inverse relation.
It does not have a constant of variation.

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Find the exponential equation for the given information. Pls help

nikitadnepr [17]

Answer:

y = 12 $(2.9)^{x}$

Step-by-step explanation:

The standard form of the exponential function is

y = a $(b)^{x}$

Find a and b by substituting ordered pairs from the table into the equation

Using (0, 12 ), then

12 = a $b^{0}$ ( $b^{0}$ = 1 ), so

a = 12, then

y = 12 $b^{x}$

Using (1, 34.8 ) , then

34.8 = 12 $b^{1}$ ( divide both sides by 12 )

b = 2.9

Thus exponential function is

y = 12 $(2.9)^{x}$

3 0

3 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0

3 years ago

Graph the following exponential functions and determine the y-intercept, domain, range, and

Ne4ueva [31]

The graph of the exponential function f(x) = 5(2)ˣ is as shown in the attached file.

<h3>How to draw the graph of an exponential Function?</h3>

We want to draw the graph of the exponential function;

f(x) = 5(2)ˣ

At input of x = 0, we have;

f(x) = 5(2)⁰ = 5

At input of x = 1, we have;

f(x) = 5(2)¹ = 10

At x = -1, we have;

f(x) = 5(2)⁻¹ = 2.5

At x = -2, we have;

f(x) = 5(2)⁻² = 1.25

Read more about Graph of Exponential Function at; brainly.com/question/12940982

#SPJ1

8 0

2 years ago

Formula of quadratic function and their graph

Mama L [17]

Answer:

-b +/- √b^2-4ac

_____________

7 0

2 years ago

Find the slope of the line that passes through (100, 56) and (28,94)

Rus_ich [418]

Answer:

-19/36

Step-by-step explanation:

We can find the slope using

m = ( y2-y1)/(x2-x1)

m = ( 94-56)/( 28 -100)

= 38 / -72

= -19/36

8 0

3 years ago