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3 years ago

The glass container shown is filled to a height of 2.25 inches. What percentage of the container is filled with sand?

Mathematics

2 answers:

m_a_m_a [10]3 years ago

4 0

1. The volume of the container shown in the figure attached, is:

V=(5 in)(4.5 in)(3 in)
V=67. 5 in³

2. The problem says that the container<span> is filled with sand to a height of 2.25 inches. Then, the volume of the sand into the container is:

V=</span>(5 in)(4.5 in)(2.25 in)
V=50.62 in³

3. Therefore, you have:

67.5 in³----100%
50.62 in³---x

x=50.62x100/67.5
x=74.99%
<span>
What percentage of the container is filled with sand?</span>

The answer is: 74.99%

dalvyx [7]3 years ago

3 0

How did u make a picture of it plz tell me

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2 years ago

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

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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.