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3 years ago

Please help, im having trouble cancelling out the numbers and letters! :) thank you

Mathematics

1 answer:

stiv31 [10]3 years ago

8 0

First change the devision to multiplication
second see the common letters and cross them

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The product of 3, and a number increased by 6, is minus−24

Andrew [12]

3(x+6) = -24

3x+18 = -24

3x = -42

x = -14

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4 years ago

Pls help me with this equation

Zanzabum

Answer:

x = 20 degree

Step-by-step explanation:

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3 years ago

Let f(x)=e^x and G(x)=x+6. What are the domain and range of g(f(x))?

Lena [83]

Answer:

Domain and Range of g(f(x)) are 'All real numbers' and {y | y>6 } respectively

Step-by-step explanation:

We have the functions, f(x) = eˣ and g(x) = x+6

So, their composition will be g(f(x)).

Then, g(f(x)) = g(eˣ) = eˣ+6

Thus, g(f(x)) = eˣ+6.

Since the domain and range of f(x) = eˣ are all real numbers and positive real numbers respectively.

Moreover, the function g(f(x)) = eˣ+6 is the function f(x) translated up by 6 units.

Hence, the domain and range of g(f(x)) are 'All real numbers' and {y | y>6 } respectively.

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3 years ago

A rubber bouncy ball is dropped from a

ss7ja [257]

Answer:

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2 years ago

Find the average value of the function g(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7Blnx%7D%7Bx%7D" id="TexFormula1" title=

aleksley [76]

Answer:

$\displaystyle average=\frac{1}{2(e-1)}$

Step-by-step explanation:

<u>Average Value of a Function</u>

Given a function g(x), we can compute the average value of g in a given interval (a,b) with the equation:

$\displaystyle average=\frac{1}{b-a}\int_{a}^{b} g(x)dx$

We use the given data

$\displaystyle average=\frac{1}{e-1}\int_{1}^{e} \frac{lnx}{x}dx$

We now compute the indefinite integral with a u-substitution

$\displaystyle I=\int \frac{lnx}{x}dx$

We'll use the substitution u=lnx, du=dx/x. Then

$\displaystyle I=\int u.du$

Integrating

$\displaystyle I=\frac{u^2}{2}$

Since u=lnx

$\displaystyle I=\frac{ln^2x}{2}$

The average value is

$\displaystyle average=\frac{1}{e-1}\left|\frac{ln^2x}{2} \right|_1^e$

$\displaystyle average=\frac{1}{e-1}\left(\frac{ln^2e}{2}-\frac{ln^21}{2} \right )$

Since lne=1, and ln1=0

$\displaystyle average=\frac{1}{e-1}\left(\frac{1}{2}-0 \right )$

$\displaystyle average=\frac{1}{2(e-1)}$

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4 years ago