Factor completely.

Given ABCE is a rectangle, D is the midpoint of CE<br> Prove AD =~ BD

Bas_tet [7]

Since the congruent operator is â‰… and since AD is congruent to BD, I'm going to assume that you want to prove that AD is congruent to BD.
1. DE is equal to CD by definition since D is the midpoint of CE.
2. AE is equal to BC since opposite sides of a rectangle are equal to each other.
3. Angle AEC is equal to Angle BCE since all angles in a rectangle are right angles and all right angles are equal to each other.
4. Triangles ADE and BDC are congruent to each other because we have SAS congruence for both triangles.
5. AD is congruent to BC since they're corresponding sides of congruent triangles.

5 0

3 years ago

A corporate attorney has weekends off and has 30 paid vacation days per year, including holidays that fall on weekdays. If his s

snow_tiger [21]

It's slightly ambiguous because the year isn't an even number of weeks, so sometimes that extra day falls on a weekend and sometimes on a weekday. We'll assume a weekend; that makes the numbers come out nice.

Total working days, 52 five day weeks minus 30 days vacation

$d = 52 \times 5 - 30 = 230$

So his daily rate is

$119600/230 = 520$ dollars per day

Answer: A

It's closer to $518 when we count the extra day as a weekday.

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.