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chubhunter [2.5K]
2 years ago
8

A, B, C y D son puntos Co lineales tal que BC es el doble de

Mathematics
1 answer:
Vanyuwa [196]2 years ago
4 0

Answer:

CD =3m

Step-by-step explanation:

AB=BD

BC=2CD --> CD=1/3 BD

--> AD= 6CD

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If m∠JKM = 43, m∠MKL = (8 - 20), and m∠JKL = (10x - 11), find each measure.
Brut [27]

Correct Question: If m∠JKM = 43, m∠MKL = (8x - 20), and m∠JKL = (10x - 11), find each measure.

1. x = ?

2. m∠MKL = ?

3. m∠JKL = ?

Answer/Step-by-step explanation:

Given:

m<JKM = 43,

m<MKL = (8x - 20),

m<JKL = (10x - 11).

Required:

1. Value of x

2. m<MKL

3. m<JKL

Solution:

1. Value of x:

m<JKL = m<MKL + m<JKM (angle addition postulate)

Therefore:

(10x - 11) = (8x - 20) + 43

Solve for x

10x - 11 = 8x - 20 + 43

10x - 11 = 8x + 23

Subtract 8x from both sides

10x - 11 - 8x = 8x + 23 - 8x

2x - 11 = 23

Add 11 to both sides

2x - 11 + 11 = 23 + 11

2x = 34

Divide both sides by 2

\frac{2x}{2} = \frac{34}{2}

x = 17

2. m<MKL = 8x - 20

Plug in the value of x

m<MKL = 8(17) - 20 = 136 - 20 = 116°

3. m<JKL = 10x - 11

m<JKL = 10(17) - 11 = 170 - 11 = 159°

3 0
3 years ago
What is 6.213 as a mixed number
Ira Lisetskai [31]
6.213 as a mixed number would be 6 and 213/1000.

all you do to convert it to a mixed number is look at how many numbers are after the decimal.

since theres 3 numbers, that means it would be in the thousands place
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Find the mode of the following data for data sets a-d (Remember what the steps are for finding the mode ).Rewrite the problem as
Nitella [24]
The mode would be 17, it’s the number that appears the most. I’m not rewriting it though
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3 years ago
NEED HELP ASAP
mezya [45]

Answer:

5, 3

Step-by-step explanation:

correct on edge2020

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2 years ago
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Solve: 4= -0.8n<br> I forgot how to do this can someone help?
frez [133]

Answer:

<em>n</em><em> </em><em>=</em><em> </em><em>-5</em>

Step-by-step explanation:

<em>Just</em><em> </em><em>divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>-0</em><em>.</em><em>8</em>

<em>:</em><em>)</em><em> </em>

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