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FromTheMoon [43]
3 years ago
11

HELP MEHHHHHHHHH. thanks.

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0
The Answer is A .....
maxonik [38]3 years ago
3 0

Answer:A

The mean is the average of all numbers, in this example it's to national averages which will have many numbers.

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Estimate the value of √15 to the nearest tenth.
Nadya [2.5K]

Answer:

3.9

Step-by-step explanation:

√15 = 3.9, rounded.

3 0
1 year ago
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Look at picture for question and answer
Sonja [21]
1. C

2. A

Hope this helped, enjoy your day.
4 0
3 years ago
Wanni cycled 6 km from her house to the school at a uniform speed, v km/h. If she increased her speed
Karolina [17]

Answer:

The quadratic equation in terms of v is   v² + 2 v + 180 = 0

Step-by-step explanation:

Given as :

The distance between house to the school = d = 6 km

The uniform speed = v km/h

So, Time = \dfrac{\textrm Distance}{\textrm speed}

or, t = \dfrac{\textrm d}{\textrm v}

Or, t = \dfrac{\textrm 6}{\textrm v}

<u>Now, Again</u>

The speed is increase by 2 km/h

i.e speed = (v + 2) km/h

So, Time taken = t' = (t - \dfrac{4}{60})hours

i.e t' =  (t - \dfrac{1}{15})hours

Now, Time = \dfrac{\textrm Distance}{\textrm speed}

So, (t - \dfrac{1}{15}) = \dfrac{\textrm d}{\textrm v}

Or,  (t - \dfrac{1}{15}) = \dfrac{\textrm 6}{\textrm (v + 2)}

Or , \dfrac{\textrm 6}{\textrm v} -  \dfrac{1}{15} = \dfrac{\textrm 6}{\textrm (v + 2)}

Or , \dfrac{\textrm 90 - v}{\textrm 15 v} = \dfrac{\textrm 6}{\textrm v + 2}

Or, (90 - v) × (v + 2) = 6 × 15 v

Or, 90 v - 180 - v² - 2 v = 90 v

Or,  v² + 2 v + 180 = 90 v - 90 v

Or,  v² + 2 v + 180 = 0

So, The quadratic equation in terms of v

v² + 2 v + 180 = 0

Hence The quadratic equation in terms of v is   v² + 2 v + 180 = 0   Answer

3 0
3 years ago
HeLPPP
sveta [45]

Answer:

1/2 is the answer. i just took the quiz

Step-by-step explanation:

7 0
3 years ago
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Whats the number of 10% of 24
Leokris [45]

Answer:

2.4

Step-by-step explanation:

10% of 24

Make the 10% a decimal.

0.10

Multiplication!!!

0.10×24

0.10×24=2.4

7 0
3 years ago
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