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SOVA2 [1]
3 years ago
11

MATH HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
sveta [45]3 years ago
6 0
So the first one, 1.2 >11/5. and the second one 1/3> 1/2
Lina20 [59]3 years ago
4 0
The first one is =, the second one is
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Suppose an airplane climbs 15 feet for every 40 feet it moves forward. What is the slope of this airplanes ascent?
Goryan [66]

Answer:

3/8

Step-by-step explanation:

Slope is the same as rise over run, or y/x. In this case, rise over run is 15 over 40, or 15/40. Simplified, this is 3/8.

8 0
3 years ago
I need helps pls :)
nadya68 [22]
In regular polygons, if the number of sides, n, is odd, the lines of symmetry will pass through a vertex and the half of the opposite side.
Step-by-step explanation:
This is equivalent to say that a perpendicular bisector of a side of the regular polygon bisects the angle of the opposite vertex to the side, and divide the polygon into two symmetrical parts.
4 0
3 years ago
Find the equation of the line shown.
vfiekz [6]

Answer:

y=x+6

Step-by-step explanation:

3 0
2 years ago
Based on given information for each of the following, which lines cannot be parallel? a m∠2=127°, m∠3=63°\ PLZZSZZZ help :>
tia_tia [17]

Answer:

Lines m and n cannot be parallel

Step-by-step explanation:

1. Because angles 2 and 3 are Same Side Interior Angles, they must be supplementary.

2. To check this you add the angles together (127 + 63) and it sums to 190 degrees, not 180

3. Since they don't agree with the laws of parallel lines, the lines that the angles are on can't be parallel

4. Angles 2 and 3 reside on lines m and n so lines m and n cannot be parallel

7 0
3 years ago
What is the area of the sector ?​
worty [1.4K]

They forget to say "not to scale".  I'm guessing this is trig because I don't see another way to do it.

Let's consider x=chord PQ first.  By the Law of Cosines

x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ

We have an isosceles triangle formed by two radii of 9 cm and x=PQ.  By the Law of Cosines again,

x^2 = 9^2 + 9^2 - 2(9)(9) \cos B

x^2 = 162 - 162 \cos B

\cos B = \dfrac{162 - x^2}{162} = 1 - \dfrac{x^2}{162}

\cos B = 1 - \dfrac{34 - 30\cos 81^\circ}{162}

B = \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ }{162} \right)

The area of the circle is the fraction given by the angle,

A = \dfrac{B}{360^\circ} \pi r ^2

A \approx \dfrac{ \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ}{162} \right) }{360} (3.142)9^2

A\approx 24.7474332960707

Answer: 24.7 sq cm

8 0
3 years ago
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