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Mathematics

1 answer:

Sedbober [7]2 years ago

8 0

Answer:

Step-by-step explanation:

x/9 = 12/18

18x = 108

x = 6

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Identify the class midpoints for the given frequency distribution.

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2 years ago

Please help!

tensa zangetsu [6.8K]

If the parabola has y = -4 at both x = 2 and x = 3, then since a parabola is symmetric, its axis of symmetry must be between x = 2 and x = 3, or at x = 5/2. Our general equation can then be:
y = a(x - 5/2)^2 + k
Substitute (1, -2): -2 = a(-3/2)^2 + k
-2 = 9a/4 + k
Substitute (2, -4): -4 = a(-1/2)^2 + k
-4 = a/4 + k
Subtracting: 2 = 2a, so a = 1. Substituting back gives k = -17/4.
So the equation is y = (x - 5/2)^2 - 17/4
Expanding: y = x^2 - 5x + 25/4 - 17/4
y = x^2 - 5x + 2 (This is the standard form.)

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2 years ago

5(x+3)−2x Which expression correctly combines the like terms? 3x+15 3x+3 3x2+15 3x2+3

arsen [322]

3x+15 , this expression correctly combines the like terms

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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.