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Nikitich [7]
3 years ago
13

If the club decides to send 5 members to the convention, how many different groups of 5 are possible? there are 10 members

Mathematics
1 answer:
hodyreva [135]3 years ago
7 0
For the 1st member, you have 10 possible choices.

For the 2nd, you have 9 remaining choices.

For the 3rd: 8.

So all possible combinations would be:
10*9*8*7*6 = 10! - 5! = 30,240 combinations.
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Find the measures of all four angles if 3·(m∠1+m∠3) = m∠2+m∠4.
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Answer:

Step-by-step explanation:

m∠1=m∠3

m∠2=m∠4

3*(m∠1+m∠3)=m∠2+m∠4

3*(m∠1+m∠1)=m∠2+m∠2

3×2 m∠1=2 m∠2

m∠2=3 m∠1

now m∠1+m∠2=180°

m∠1+3 m ∠1=180

4 m∠1=180

m∠1=180/4=45°

m∠3=45°

m∠2=180-m∠1=180-45=135°

m∠4=135°

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3 years ago
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Explain What Happens To The Volume Of A Sphere When
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Find the median of these numbers:4,2,7,4,3 *
sattari [20]

Answer:

The Median of those numbers is 4

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1. Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer t
Mazyrski [523]

Answer:

Part (A): The required inequality is T < 35 or t >40.

Part (B): The correct option is C) Only x=7.

Part (C) |x+1|+5=2 has no solution; |4x+12|=0 has one solution; |3x|=9 has two solution.

Step-by-step explanation:

Consider the provided information.

Part (A)

Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer than 40°F is an unsafe food practice.

The union is written as A∪B or “A or B”.

The intersection of two sets is written as A∩B or “A and B”

We need to determine the inequalities represent the union of these improper storage.

That means we will use A∪B or “A or B”.

The improper storage temperatures is when temperature is less than 35°F or greater than 40°F.

Hence, the required inequality is T < 35 or t >40.

Part (B) 15| x-7|+4=10|x-7|+4

Solve the inequality as shown below:

Subtract 4 from both sides.

15| x-7|+4-4=10|x-7|+4-4

Subtract 10|x-7| from both sides

15\left|x-7\right|-10\left|x-7\right|=10\left|x-7\right|-10\left|x-7\right|

5\left|x-7\right|=0\\\left|x-7\right|=0\\x=7

Hence, the correct option is C) Only x=7.

Part (C) Match the solution,

\left|x+1\right|+5=2

Subtract 2 from both sides.

\left|x+1\right|=-3

Absolute value cannot be less than 0.

Hence, |x+1|+5=2 has no solution.

|4x+12|=0

4x+12=0

x=-3

Hence, |4x+12|=0 has one solution.

|3x|=9

\mathrm{If}\:|u|\:=\:a,\:a>0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a

3x=-9\quad \mathrm{or}\quad \:3x=9\\x=-3\quad \mathrm{or}\quad \:x=3

Hence, |3x|=9 has two solution.

3 0
3 years ago
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