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kompoz [17]
3 years ago
14

In the diagram, if KL=10, and MK=2, and JM=6 determine the value of MN. Show your work or explain how you arrived at your answer

.
Please help

Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0

Answer:

6

Step-by-step explanation:

JM is 6 riiite

and JMN should be an equilateral triangle so

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Y = x – 6 x = –4 what is the solution to the system of equations? (–8, –4) (–4, –8) (–4, 4) (4, –4
fredd [130]

Answer:

(- 4, - 10 )

Step-by-step explanation:

Given the 2 equations

y = x - 6 → (1)

x = - 4 → (2)

Substitute x = - 4 into (1) for corresponding value of y

y = - 4 - 6 = - 10

Solution is (- 4, - 10 )

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What is the perimeter of this triangle
DENIUS [597]

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they are all equal side so add them up

Step-by-step explanation:

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Ben is paid £8.50 per hour.
skelet666 [1.2K]

Answer:

£103.70

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3 0
3 years ago
Which applies the power of a product rule to simplify (5t)^3?
frozen [14]
(5t)^3 = 125t^3
5 0
3 years ago
Read 2 more answers
Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow
prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

CI=\hat{Y}\pm t_{Critical}\times S_{e}

Where,

\hat{Y}= predicted value of lead content when traffic flow is 15.

\to df=n-1=8-1=7

 95\% \ CI\  is\  (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\

     =\frac{1059.8}{2}\\\\=529.9

Calculating thet-critical valuet_{ \{\frac{\alpha}{2},\ df \}}=-2.365

The lower predicted value =529.9-2.365(Se)

463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076

When 99\% of CI use as the expected lead content: \to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)

8 0
3 years ago
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