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3 years ago

Help this is due today!

Mathematics

1 answer:

Luda [366]3 years ago

8 0

Answer:

Step-by-step explanation:

Automatically if you use the intercept (where it cuts the line) which would be 3.

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Donald says he uses a standard number cube for simulations that involve 2, 3, or 6 equal outcomes. Complete the explanation for

Alexxandr [17]

Answer:

Step-by-step explanation:

7 0

3 years ago

EXPLANATION?????

crimeas [40]

Answer:

Hello! answer: x = 19

Step-by-step explanation:

This is a supplementary angle! So it will add up to 180 degrees 180 - 123 = 57 so we know something multiplied by 3 will equal 57 and 19 × 3 = 57 so x = 19 Hope this is a good explanation!

4 0

3 years ago

Read 2 more answers

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

4 years ago

An exponential function is expressed in the form y=axb^x. The relation represents a growth when _______ and a decay when ______.

seropon [69]

Answer:

Growth when: b>1.

Decay when: 0<b<1.

Step-by-step explanation:

Any function in the form $f(x)=a{\times}b^x$ , where a > 0, b > 0 and b not equal to $1$ is called an exponential function with base b.

If 0 < b < 1 this is an example of an exponential decay.

The general shape of an exponential with b > 1 is an example of exponential growth.

Hence,

An exponential function is expressed in the form $f(x)=a{\times}b^x$ , The relation represents a growth when b >1 and a decay when 0<b<1.

5 0

3 years ago

W = 3x + 7y solve for y

mario62 [17]

Answer:

The value of the equation $y=\frac{W-3x}{7}$ .

Step-by-step explanation:

Consider the provided equation.

$W = 3x + 7y$

We need to solve the provided equation for y.

Subtract 3x from both side.

$W-3x= 3x-3x+ 7y$

$W-3x=7y$

Divide both sides by 7.

$\frac{7y}{7}=\frac{W-3x}{7}$

$y=\frac{W-3x}{7}$

Hence, the value of the equation is $y=\frac{W-3x}{7}$ .

8 0

3 years ago