What is the length of Line segment A C? Round to the nearest tenth. 3.0 cm. 9.8 cm. 10.5 cm. 12.8 cm.

Mathematics

2 answers:

12345 [234]2 years ago

6 0

Answer:

10.5

Step-by-step explanation:

Simora [160]2 years ago

5 0

Answer:

10.5 cm

Step-by-step explanation:

A line segment is defined as a part of a line. It is a straight line having two end points. The only difference between a line and line segment is that a line is continuous with indefinite length whereas a line segment has a definite length to it between the two points.

In the figure, the length of the line segment is 10.5 cm (rounded to the nearest tenth).

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Given: 315/1575. Write the prime fractorization of the numerator and the denominator, then write the fraction in lowest terms.

Ne4ueva [31]

The prime factorization of the numerator, 315, is 5 x 3 x 3 x 7. That of the denominator, 1575 is 5 x 5 x 3 x 3 x 7. To convert the fraction to its lowest form, cancel all the factors that appear to both the numerator and denominator. We will be left with 1/5.

3 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.