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Naddik [55]
3 years ago
7

How is a slant asymptote present in the equation f(x)= x/(x-1)

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
5 0
Hello here is a solution :

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PLS HELP I WILL REALLY APPRECIATE PLS I WILL MARK YOU THE BRAINLY !
otez555 [7]

Answer:

try with this link https://es.symbolab.com/solver/square-roots-calculator

Step-by-step explanation:

3 0
3 years ago
Write a system of inequalities that defines a shaded region that looks like a<br> right triangle.
eduard

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

<h2>Explanation:</h2>

A right triangle is a special triangle that has a right angle. In this case, we have to write a system of inequalities that defines a shaded region that looks like a  right triangle. First of all, let's say that at the origin the triangle will have a right angle. To do so, we'd need to set:

x>0 \\ \\ y>0

So the shaded region in this first part will be the First Quadrant as indicated in the first figure below. So if the opposite side lies on the x-axis the adjacent side will lie on the y-axis or if the adjacent side lies on the x-axis the opposite side will lie on the y-axis. Everything is ok up to this point. We just need to define the hypotenuse, so we'd need to define a line.  In order to have a right triangle, we need a line with negative slope and positive y-intercept shaded under the line. So, this inequality could be:

y

Finally, the system of inequalities would be:

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

And the final shaded region is the one shown in the second figure.

<h2>Learn more:</h2>

Inequalities: brainly.com/question/2486051

#LearnWithBrainly

3 0
3 years ago
What is the slope of the line that passes through the points (-2, -5)(−2,−5) and (-3, -6) ?(−3,−6)? Write your answer in simples
pogonyaev

Answer:

0 & 0

Step-by-step explanation:

In order to find the answer to these two questions you need to use the slope formula, substitute, subtract, and simplify if needed.

(-2,-5),(-2,-5) = (x^1,y^1),(x^2,y^2)

Slope formula = m=\frac{y^2-y^1}{x^2-x^1}

Substitute: \frac{-5--5}{-2--2} =

\frac{-5--5}{-2--2} =-5--5 = 0

-2--2 = -2+2=0

= 0

(-3,-6),(-3,-6)=(x^1,y^1),(x^2,y^2)

m=\frac{-6--6}{3--3}

-6+6=0

-3+3=0

=0

Hope this helps.

7 0
2 years ago
Find the<br>Prove that sin theta tan theta= (1 + sec theta)(1 - cos theta)<br>​
larisa86 [58]

Answer:

Step-by-step explanation:

RHS = (1 +Sec Ф)(1 - Cos Ф)

       = 1*1 -  1*Cos Ф + Sec Ф *1 - Sec Ф *Cos Ф

      = 1 - Cos Ф + Sec Ф - 1              {Sec Ф = \frac{1}{Cos \ theta} )

       =Sec Ф - Cos Ф

       = \frac{1}{Cos \ theta} - Cos Ф

       = \frac{1}{Cos \ theta}-\frac{Cos \ theta*Cos \ theta}{Cos \ theta}\\\\= \frac{1-Cos^{2} \ theta}{Cos \ theta}\\\\= \frac{Sin^{2} \ theta}{Cos \ theta}\\\\=\frac{Sin \ theta*Sin \ theta}{Cos \ theta}\\\\= Sin \ theta*\frac{Sin \ theta}{Cos \ theta}

        = Sin  Ф* tan  Ф = LHS

5 0
3 years ago
Read 2 more answers
(-2x²y)³ / (xy²z)² * (-yz)² Hello guys, im having trouble with this babe...
Masja [62]
\frac{(-2x^2y)^3}{(xy^2z)^2(-yz)^2}

On top: (-2)³ = -2 × -2 × -2 = -8
(x²)³ = x^6 (the exponents multiply)
and of course, (y)³ = y³

On the bottom: (xy²z)² = x² y^4 z²
(-yz)² = y²z²
Multiplying these together, the exponents add and we get x² y^6 z^4.

\frac{-8x^6y^3}{x^2y^6z^4}

So, your reasoning is correct for what you have so far.

Your next step would be cancelling shared factors from the top and bottom.
Just like with regular fractions, if the numerator and denominator are divisible by the same number, you can divide them by it to simplify. (ex: 4/6 = 2/3)

Well, x^6 and x^2 are both divisible by x^2, right?
We can also cancel the y^3.

It might help to visualise the factors like this:

\frac{-8xxxxxxyyy}{xxyyyyyyzzzz}

Once you've cancelled out x² and y³ from each, you're left with

\boxed{\frac{-8x^4}{y^3z^4}}
7 0
4 years ago
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