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Rudiy27
3 years ago
6

Which conic section does the equation below describe?

Mathematics
1 answer:
wel3 years ago
5 0
Answer is circle


Canonical equation for circle is (x — x0)2 + (3, yo)2 = R2 ,
hence (x + 1)2 + (y — 3)2 = 4 describes a circle.
Answer: C Circle.
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One zero of x^3 - 4x = 0 is 0 what are the other zeros of the function
myrzilka [38]
So you have x^3 - 4x = 0. What you can do is pull out an x from both x^3 and - 4x so it looks like this:

x( {x}^{2} - 4) = 0

Then you can find a number that makes the part inside the parentheses turn into zero. For beginners, it may be easier to write it out seperately and solve for x.

{x}^{2} - 4 = 0

We need to solve for x, so the first step is to add 4 to both sides, so we get something like this:

{x}^{2} = 4

Then, we can square root both sides to get rid of the power on the x, so it looks like this:

x = \sqrt{4}

Now, every square root has two answers, a positive and a negative. If we look at the bottom example:

{2}^{2} = 4

{( - 2)}^{2} = 4

We can see that both -2 and 2 to the power of two will equal to 4.

So finally, we get:

x = - 2 \: and \: 2

These are the other 'Zero's for the original function. If you are not sure of what a 'Zero' is, it is where the function crosses over the x-axis on a graph.
5 0
3 years ago
Which is equal to -214°?
Neporo4naja [7]
Answer: -107pi/45 radians
5 0
3 years ago
Liliana wants to write an equivalent expression for n+3-5n+6.
sammy [17]

Answer:

Step-by-step explanation:

n +3 - 5n + 6 = n - 5n + 3 +6

= -4n + 9

4 0
3 years ago
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4 years ago
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Find all solutions of the given system of equations (If the system is infinite many solution, express your answer in terms of x)
lisov135 [29]

Answer:

(a) The system of the equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. has no solution.

(b) The system of the equations \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. has many solutions y=\frac{2x}{3}-\frac{5}{3}

Step-by-step explanation:

(a) To find the solutions of the following system of equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. you must:

Multiply 2x-3y=3 by 2:

\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}

Subtract the equations

4x-6y=3\\-\\4x-6y=6\\------\\0=-3

0 = -3 is false, therefore the system of the equations has no solution.

(b) To find the solutions of the system \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. you must:

Isolate x for 4x-6y=10

x=\frac{5+3y}{2}

Substitute x=\frac{5+3y}{2} into the second equation

16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40

The system has many solutions.

Isolate y for 4x-6y=10

y=\frac{2x}{3}-\frac{5}{3}

3 0
3 years ago
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