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3 years ago

Please help I will give brainlist

Mathematics

2 answers:

Alex777 [14]3 years ago

8 0

Answer: Option B (y=8x+14)

Step-by-step explanation:

To solve these kind of question we just simply keep on inserting the values of x and if we get the required value of y that is in the table that equation is our answer, For example lets take Option A,

y = 8x

If we insert x = 0 we get,

y = 8(0)

y=0

but in the table it says we get 14 so option A is incorrect

Like this we keep checking for more values and all the values of x satisfy our equation in Option B like this,

y=8x+14

y=8(0)+14

y=14

y=8x+14

y=8(1)+14

y=8+14

y=22

y=8x+14

y=8(2)+14

y=16+14

y=30

and so on. Now your answer would be Option B.

THE CREDIT GOES TO (IjlalHashmi)

Mariulka [41]3 years ago

4 0

Answer:

Your answer will be B

Step-by-step explanation:

Because you find the difference between 14 and 22 or the difference between 22 and 30 or the difference between 30 and 38. and so on then it will be 8

You then say ok 8 times x so for example 8×1 = 8 then you say ok 22 - 8= 22. So it will be 8x+14...

I hope this helped if not tell me please....❤

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

zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

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

2) if its vertex is below the x-axis, it is 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 not have 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 $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.