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Lilit [14]

9514 1404 393

Answer:

x = 30
y = 105
z = 75

Step-by-step explanation:

You know the linear pair z° and 105° are supplementary angles, so ...

z = 180 -105 = 75

The other base angle of the isosceles triangle has the same measure, 75°. __

Then x can be found either from the sum of interior angles of the triangle, or from the relation of 105° to the "remote interior angles". The first relation gives ...

75° +75° +x° = 180° ⇒ x = 180 -150 = 30

The second relation gives ...

75° +x° = 105° ⇒ x = 105 -75 = 30

__

y° is supplementary to the left-side base angle, so is ...

y = 180 -75 = 105

Of course, you could also figure y from the symmetry of the figure.

The values of x, y, z are 30, 105, 75, respectively.

5 0

3 years ago

PLEASE HELP, I WILL UP POINTS AND VOTE BRAINLIEST! DONT MAKE ME WASTE MY 88 POINTS.

Allushta [10]

Answer:

Ok im using a standered form calculator:

1. 9/7x + 1/7 = -8

2. y = -6/5x - 3 = -11

3. y = 1/3x - 2 = -10

Step-by-step explanation:

I will be glad to help more

6 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Mr. Jones used a fair spinner divided into 4 sections as shown below. They spun the spinner 50 times and recorded the results in

tatuchka [14]

Answer:

The theoretical probability would be 25% or 1/4 because there are four sections but in the example the probability of a one being spun was 24%

Step-by-step explanation:

4 0

2 years ago

Which of the following relations has a domain of {2, 3, 6}? Choose all that apply. {(3, 3), (2, 2), (3, 2), (6, 1)} {(3, 1), (6,

Ganezh [65]

All except (0,2), (5,6), (5,3), and (4,3). The domain is the x and the range is the y, so the x coordinate has to either be 2, 3, or 6.

7 0

2 years ago