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katrin [286]
3 years ago
5

Find the value of x. (6x + 1) (9x - 23)

Mathematics
1 answer:
shtirl [24]3 years ago
3 0

Answer:

Step-by-step explanation:

(6x+1)(9x-23)= 6x(9x-23)+1(9x-23)

= 54x²-138+9x-23

= 54x²+9x-161

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Which quadratic equation fits the data in the table?
marissa [1.9K]

Answer:

y = 3 - x + x²

Step-by-step explanation:

Given the data:

x. y

-5 33

-2 9

-1 5

0 3

3 9

4 15

6 33

General formof a quadratic model:

y = A + Bx + Cx²

Using the quadratic regression model solver for the data Given:

The quadratic model fit obtained is :

y = 3 - x + x²

4 0
3 years ago
Please help me ASAP <br> NO LINKS or files
Bess [88]

Answer:

1 (-9, -4)

2 (-6, 2)

Step-by-step explanation:

1- when you reflect along the x axis keep the x the same and change the y.

2- when you reflect along the y axis keep the y the same and change the x.

4 0
3 years ago
What is the result to 4.4a-7.3
Ymorist [56]
4.4a-7.3 is equal to
4.4a= 7.3 divide both sides by 4.4
a= 1.66
6 0
3 years ago
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.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
Bike rental company A charges an initial fee of $8 plus $1.25 for every hour spent biking. Bike rental company B charges $2.75 f
Lera25 [3.4K]
2.75h =8-1.25h is thus the one
7 0
3 years ago
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