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Dmitriy789 [7]
3 years ago
10

Identify the constant term in the function. y=-x^2+3x-4

Mathematics
1 answer:
Alexeev081 [22]3 years ago
3 0
The constant term is -4
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a parabola has an x-intercept at 2, its axis of symmetry is the line x=4, and the y-coordinate of its vertex is 6. Determine the
nlexa [21]

Answer:

The standard equation of the parabola is:

y=-\frac{3}{2}x^2+12x-18

Step-by-step explanation:

An x intercept of 2 means that the point (2, 0) is in the graph of the parabola.

We can also write the general expression for the parabola in vertex form, since we can use the information on the coordinates of the vertex: (4, 6) - recall that the axis of symmetry of the parabola goes through the parabola's vertex, so the x-value of the vertex must be x=4.

y-y_{vertex}=a\,(x-x_{vertex})^2\\y-6=a\,(x-4)^2

Now we can find the value of the parameter "a" by using the extra information about the point (2, 0) at which the parabola intercepts the x-axis:

y-6=a\,(x-4)^2\\0-6=a\,(2-4)^2\\-6=a\,4\\a=-\frac{6}{4} =-\frac{3}{2}

Then the equation of the parabola becomes:

y-6=-\frac{3}{2} \,(x-4)^2\\y-6=-\frac{3}{2} (x^2-8x+16)\\y-6=-\frac{3}{2}x^2+12x-24\\y=-\frac{3}{2}x^2+12x-18

7 0
3 years ago
a circle has a diameter of 10cm what is the best proximation of this area use 3.14 to approximate for pi
hammer [34]
Hello!

The answer would be 75.8^{cm2}

Hope this helped!
4 0
3 years ago
Read 2 more answers
Select all the numbers that are common multiples of 4 and 6.
Verdich [7]

Answer:

12, 24, and 60.

Step-by-step explanation:

It says multiples. Not factors, so it can't be 1 or 2. Neither of the numbers can make the number 10. 4 can make 40, but 6 can't.

8 0
3 years ago
Read 2 more answers
Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1
sasho [114]
I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse. 

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.) 

\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1

Where (x_{1},y_{1}) is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is: 

(-3,5)

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis. 

The major axis in this case is that of the y-axis. In other words, 

b^2>a^2 

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are: 

(-3,10)
 & (-3,0)

I really hope this helped you! (Partially because I spent a lot of time on this lol) 

Sincerely,

~Cam943, Junior Moderator
6 0
4 years ago
-10x-6y=0 -4x -6y=54 find the solution of this system of equations
ad-work [718]

You can see that the term -6y appears in both equations. In this cases, we can leverage this peculiarity and subtract the two equations to get rid of the repeated term. So, if we subtract the first equation from the second, we have

(-4x -6y) - (-10x-6y) = 54-0 \iff 6x = 54 \iff x = 9

Now that we know the value of x, we can substitute in any of the equation to deduce the value of y: if we use the first equation, for example, we have

-10x-6y=0 \iff -10\cdot 9 - 6y = 0 \iff -90-6y=0 \iff 6y = -90 \iff y = -15

7 0
3 years ago
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