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3 years ago

Pls help with 2 and 3

Mathematics

1 answer:

pychu [463]3 years ago

5 0

Answer:

you determine based on the x and y value (x,y) based on the numbers you place it in the corresponding quadrant based on negatives and positives then plot the point on the graph

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-3=3x-3.solve for x?

Sunny_sXe [5.5K]

Add 3 on both sides of the equation

3x=0

divide 3 on both sides of the equation

x=0

7 0

3 years ago

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Find the solution of the system of equations. −5x+5y=-10 6x+5y=1

pickupchik [31]

Answer:

Step-by-step explanation:

Since you have

in one equation and

−

in the other equation, you can add both equations to cancel out the y terms and solve for x.

−

therefore

−

multiplying both sides by -1:

−

plugging this back into the first equation:

−

(

−

)

subtracting 66 from both sides:

−

divide both sides by 5:

−

putting the x-values and y-values into one point gives:

(

−

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as the solution

4 0

3 years ago

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Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

3 0

3 years ago

Please view the picture for the problem.

Tema [17]

Notice that we have common denominator 2, so we can split the denominator to express the function in the form $f(x)=mx+b$
where
$m$ is the slope
$b$ is the y-intercept

$f(x)= \frac{3x-1}{2}$
$f(x)= \frac{3}{2} x- \frac{1}{2}$
Look what we have here, a linear function! We know that the domain and range of linear functions is the real numbers.
We can conclude that:
the domain of the function is all the real numbers
the range of the function is all the real numbers

Now, to find the inverse of our function, we are going to replace $f(x)$ with $y$ ; then, we are going to interchange $x$ and $y$ and solve for $y$ :
$f(x)= \frac{3}{2} x- \frac{1}{2}$
$y= \frac{3}{2} x- \frac{1}{2}$
$x=\frac{3}{2} y- \frac{1}{2}$
$\frac{3}{2} y=x+ \frac{1}{2}$
$3y=2x+1$
$y= \frac{2}{3} x+ \frac{1}{3}$
$f^{-1}(x)=\frac{2}{3} x+ \frac{1}{3}$
The inverse of our linear function, is another linear function, so we can conclude that:
the domain of the inverse function is all the real numbers
the range of the inverse function is all the real numbers

4 0

3 years ago

Which set of points is a function

MArishka [77]

What are the sets of points?

3 0

3 years ago