Two angles of a triangle measure 22° and 530. What is the measure of the
third angle?

Answer:  T=1700

Step-by-step explanation:

Answer:

Directions: Write the new coordinates for each point or figure.

Sample Question:

Point F has the coordinates (-9,6). What are the coordinates of its image point after the translation of six units down and three units to the left?

Answer:

The coordinates of its image point after the translation are (-12, 0).

Explanation: When you look at an ordered pair the x number comes first, followed by the y number (x,y). Moving a point left or right changes the x number. Moving a point up or down changes the y number. Left and down require subtraction, while right and up require addition. In the sample, we had to go 3 units to the left, -9 - 3 = -12. We also had to go 6 units down, so 6 - 6 = 0.

Point Q has the coordinates (9,5). What are the coordinates of its image point after a translation of six units up and three units to the right?

Point P has the coordinates (3,4). What are the coordinates of its image point after a translation of 8 units down and 5 units to the left?

Point Z has the coordinates (10,7). What are the coordinates of its image point after a translation of six units down and three units to the right?

Point Y has the coordinates (0,-4). What are the coordinates of its image point after a translation of 10 units up and 3 units to the left?

Point X has the coordinates (3,-8). What are the coordinates of its image points after a translation of 11 units down and 10 units to the right?

Answer:

(-10,5)

Step-by-step explanation:

Answer:

[tex] x_{1} = 3 + \sqrt {6} [/tex]

[tex] x_{2} = 3 - \sqrt {6} [/tex]

Step-by-step explanation:

Given the quadratic equation;

x² - 6x + 3 = 0

To find the roots of the quadratic equation, we would use the quadratic formula;

Note: the standard form of a quadratic equation is ax² + bx + c = 0

a = 1, b = -6 and c = 3

The quadratic equation formula is;

[tex] x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a} [/tex]

Substituting into the formula, we have;

[tex] x = \frac {-(-6) \; \pm \sqrt {-6^{2} - 4*1*(3)}}{2*1} [/tex]

[tex] x = \frac {6 \pm \sqrt {36 - (12)}}{2} [/tex]

[tex] x = \frac {6 \pm \sqrt {36 - 12}}{2} [/tex]

[tex] x = \frac {6 \pm \sqrt {24}}{2} [/tex]

[tex] x = \frac {6 \pm 2 \sqrt {6}}{2} [/tex]

[tex] x_{1} = \frac {6 + 2 \sqrt {6}}{2} [/tex]

[tex] x_{1} = \frac {6}{2} + \frac {2 \sqrt {6}}{2} [/tex]

[tex] x_{1} = 3 + \sqrt {6} [/tex]

Or

[tex] x_{2} = \frac {6 - 2 \sqrt {6}}{2} [/tex]

[tex] x_{2} = \frac {6}{2} - \frac {2 \sqrt {6}}{2} [/tex]

[tex] x_{2} = 3 - \sqrt {6} [/tex]