16 POINTS. Find x and y intercepts and show work. I’m looking for number six.

Mathematics

1 answer:

KonstantinChe [14]3 years ago

5 0

What an awful question.
y = (x^4 - 3x^3 + 4x)^2
y = x^2(x^3 - 3x^2 + 4)^2
Go to Wolframalpha and put the above equation into the input line.
You see that there are 2 solutions at -1
There are 2 solutions at 0
There are 2 solutions at 2
You know that there are 2 solutions for each because the graph just touches the x axis. I wouldn't both to try and solve this thing using algebra.

that's what the graph would leave you to believe. There are 2 other solutions and they must be imaginary.

Note to moderator. Wolfram dot com just shows graphs. the work is mine and I'm just using a tool to explain my answer.

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I don’t understand this.. <br> will someone please help? Will mark Brainliest along with 20 points.

inn [45]

To say two quantities $a$ and $b$ "vary directly" with one another means that a change in one of the quantities results in a proportional change in the other quantity. For example, if $a=2b$ , then $b$ is proportional to $a$ by a factor of 2, or $a$ is proportional to $b$ by a factor of 1/2. In other words, if you change $b$ by some fixed amount, then $a$ changes by double that amount.

Mathematically this means there is some constant scaling factor $k$ such that $a=kb$ . In question 4, we're told $f(x)$ varies directly with $x^2$ , which means there's some $k$ such that

$f(x)=kx^2$

We're given that $f(x)=40$ when $x=2$ , so

$40=k\cdot2^2\implies40=4k\implies k=10$

Then when $x=5$ , we get

$f(5)=10\cdot5^2=10\cdot25=250$

- - -

For $a,b$ to "vary inversely" with one another means that a change in one quantity results in an "opposite" change in the other. In other words, if $a$ increases, then $b$ decreases, and vice versa.

Mathematically, this is the same as saying there is some fixed number $k$ such that $ab=k$ .

For example, if $k=1$ and we start with $a=1$ , then $b=1$ also. If we change to $a=2$ , then $b=\dfrac12$ ; if $a=10$ , then $b=\dfrac1{10}$ , and so on.