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Aliun [14]
3 years ago
9

HELPPP ILL MARK BRAINLIEST

Mathematics
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

m\angle MON = 15^\circ

Step-by-step explanation:

Thanks!

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In the parallelogram above, BE= 7x-12, CE=10x-5, and DE= 3x+4. What is the length of BD​
kompoz [17]

ABCD is a parallelogram ⇒ BE = DE

So: 7x - 12 = 3x + 4

⇔ 7x - 3x = 4 + 12

⇔ 4x = 16

⇔ x = 16/4 = 4

⇒ BE = 7(4) - 12 = 28 - 12 = 16

⇒ DE = 3(4) + 4 = 12 + 4 = 16

⇒ BD = BE + DE = 16 + 16 = 32

Answer: BD = 32

Ok done. Thank to me :>

3 0
2 years ago
Heather’s puppy weighs 23 pounds he has been gaining 3 pounds every month as he grows if this pattern continues how much will th
miskamm [114]
If the puppy weighs 3 pounds every month for five months, (5 x 3), that means he will gain 15 pounds in 5 months. Ad when you add that the original weight, (23 + 15), the puppy will weigh 38 pounds  five months from now. 
7 0
3 years ago
Read 2 more answers
Help a bab out please?
Over [174]

Answer: 5

Step-by-step explanation:

7 0
3 years ago
-36x400+931/10000<br> Please help!
AlekseyPX

Answer:

-14399.9069

Step-by-step explanation:

6 0
3 years ago
Q 2 PLEASE HELP ME FIGURE THIS OUT
suter [353]

Answer: IV, positive, \frac{\pi} {6}, - sec \frac{\pi} {6}, \frac{2\sqrt{3}}{3}

<u>Step-by-step explanation:</u>

a) Look at the Unit Circle to see that \frac{11\pi} {6} = 330°, which is located in Quadrant IV.

b) The coordinate (cos θ, sin θ) for \frac{11\pi} {6} is: (\frac{\sqrt{3}} {2},\frac{-1}{2})

sec = \frac{1}{cos} = \frac{2}{\sqrt{3}} which is positive

c) Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the given angle \frac{11\pi} {6} from 2π: \frac{12\pi} {6} - \frac{11\pi} {6} = \frac{\pi} {6}

d) the reference angle is below the x-axis so the given angle is equal to the negative of the reference angle: - sec \frac{\pi} {6}.

e) sec \frac{11\pi} {6} = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

***************************************************************************************

Answer: \frac{18\pi}{11}, IV, \frac{4\pi} {11}

<u>Step-by-step explanation:</u>

2π is one rotation.  2π = \frac{22\pi}{11}

\frac{-26\pi}{11} + \frac{22\pi}{11} = \frac{-4\pi}{11}

\frac{-4\pi}{11} + \frac{22\pi}{11} = \frac{18\pi}{11}

Convert the radians into degrees to see which Quadrant it is in by setting up the proportion and cross multiplying:

\frac{\pi}{180}= \frac{18\pi}{11x}

π(11x) = (180)18π

x = \frac{180(18\pi}{11\pi}

x = 295°     <em>which lies in Quadrant IV</em>

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value\frac{18\pi} {11} from 2π: \frac{22\pi} {11} - \frac{18\pi} {11} = \frac{4\pi} {11}

***************************************************************************************

Answer: \frac{5\pi}{3}, IV, \frac{4\pi} {11}, \frac{\pi} {3}

<u>Step-by-step explanation:</u>

2π is one rotation.  2π = \frac{6\pi}{3}

\frac{-13\pi}{3} + \frac{6\pi}{3} = \frac{-7\pi}{3}

\frac{-7\pi}{3} + \frac{6\pi}{3} = \frac{-\pi}{3}

\frac{-\pi}{3} + \frac{6\pi}{3} = \frac{5\pi}{3}

This is on the Unit Circle at 300°, which is located in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value\frac{5\pi} {3} from 2π: \frac{6\pi} {3} - \frac{5\pi} {3} = \frac{\pi} {3}


7 0
4 years ago
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