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Ivan
3 years ago
7

Label each angle with degrees and radians. label all 4 quadrants

Mathematics
1 answer:
Andru [333]3 years ago
4 0
I need a answer ASAP let me know when you guys get it please
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A system of two linear inequalities is graphed as shown, where the solution region is shaded. Complete the sentences below by de
andrezito [222]

All the points  The point (-3.2), The point (-7.5), Point (2.-5), and The point (-14,6) is located in the solution region.

<h3>What is a graph?</h3>

A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.

When we plot all the points on the graph we can see that all the points The point (-3.2), The point (-7.5), Point (2.-5), and The point (-14,6) are located in the solution region.

Therefore all the points  The point (-3.2), The point (-7.5), Point (2.-5), and The point (-14,6) is located in the solution region.

To know more about graphs follow

brainly.com/question/25020119

#SPJ1

6 0
2 years ago
9x+1/2y51<br> 7x+1/3y=39
olga2289 [7]
What’s the question though lol
7 0
2 years ago
F(X)=2x^2 +1 find and simplify:: if f(a) =19 find a.<br> Isn't a just 19??
liberstina [14]
2a^2+1=19\\&#10;2a^2=18\\&#10;a^2=9\\&#10;a=-3 \vee a=3
5 0
3 years ago
Given that A, O &amp; B lie on a straight line segment, evaluate obtuse ∠AOC.
Evgesh-ka [11]

Answer:

AOC = 124°

Step-by-step explanation:

Angle on a straight line is 180°, therefore, the sum of the three angles shown is 180;

This can be written like so:

(3x + 94) + (x + 30) + (2x - 4) = 180

This equation can be solved to find x:

6x + 120 = 180

6x = 180 - 120

6x = 60

x = 10

AOC = 3(10) + 94

= 30 + 94

= 124

7 0
2 years ago
Read 2 more answers
Given: <br> ΔABC, m∠C = 90º, m∠B = 30º<br> CM - angle bisector<br><br> Find: m∠AMC
SashulF [63]

Answer:

75°

Step-by-step explanation:

1. Consider right triangle ABC. In this triangle

m∠C = 90º and CM - angle bisector.

If CM is angle C bisector, then

m∠ACM = m∠MCB = 45º

2. Consider triangle CMB. The sum of the measures of all interior angles of the triangle is always 180°, then

m∠MCB + m∠CBM + m∠BMC = 180°,

m∠BMC = 180° - 45° - 30°

m∠BMC = 105°

3. Angles AMC and BMC are supplementary angles, so

m∠AMC + m∠BMC = 180°

m∠AMC = 180° - 105°

m∠AMC = 75°

3 0
3 years ago
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