What’s the difference from linear and non linear

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

4 0

3 years ago

Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit

Virty [35]

Hence it is a simply rearrangement of the equation to start with, in order to make  the subject:

This is the graph in 'slope-intercept' form. From here it is easy to see that gradient = and that y-intercept = 490.

The easiest way to draw a straight-line graph, such as this one, is to plot the y-intercept, in this case (0, 490), then plot another point either side of it at a fair distance (for example substitute  = -5 and  = 5 to procure two more sets of co-ordinates). These can be joined up with a straight line to form a section of the graph, which would otherwise extend infinitely either side - use the specified range in the question for x-values, and do not exceed it (clearly here the limit of -values is 0 ≤ x ≤ 735, since neither x nor y can be negative within the context of the question - the upper limit was found by substituting  = 0).

In function notation, the graph is:

The graph of this function represents how the value of the function varies as the value of x varies. Looking back at the question context, this graph specifically represents how many wraps could have been sold at each number of sandwich sales, in order to maintain the same profit of $1470.

When the profit is higher, the gradient is not changed (this is defined by the relationship between the $2 and $3 prices, not the overall profit) - instead the -intercept is higher:

Therefore we have gleaned that the new y-intercept is.

Clearly I cannot see the third straight line. However the method for finding the equation of a straight line graph is fairly simple:

1. Select two points on the line and write down their coordinates
2. The gradient of the line =
3. Find the change in  (Δ
4. Find the change in  (Δ
5. Divide the result of stage 3 by the result of stage 4
6. This is your gradient
7. Take one of your sets of coordinates, and arrange them in the form , where your  is the gradient you just calculated
8. There is only one variable left, which is  (the y-intercept). Simply solve for this
9. Now generalise the equation, in the form , by inputting your gradient and y-intercept whilst leaving the coordinates as  and

For example if the two points were (1, 9) and (4, 6):

Δ = 6 - 9 = -3
Δ = 4 - 1 = 3
=  = -1
I choose the point (4, 6)
6 = (-1 * 4) + c
6 = c - 4
c = 10
Therefore, generally,

Within the context of the question, I imagine the prices of the two lunch specials will be the same in the third month and hence the gradient will still be  - this means steps 1-6 can be omitted. Furthermore if the axes are clearly labelled, you may even be able to just read off the y-intercept and hence dispose with steps 1-8!

5 0

3 years ago

A circle has an arc of length 24π that is intercepted by a central angle of 80°. What is the radius of the circle?

bixtya [17]

I got radius= 54
arc/2pi r = degrees/360
then fill in what you know: 24pi/2pi r = 80/360
Now cross multiply and you get: 8640pi = 160pi r
Now divide both sides by 160pi and you get: 54= r

8 0

3 years ago

Write the equation of the line in fully simplified slope-intercept form?

Vinvika [58]

Answer:

y= -2x+1

Step-by-step explanation:

i hope this helps :)

7 0

3 years ago

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Compute.<br> 32 · 43<br> what is the anser