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amid [387]
3 years ago
15

are given the sets A = (even numbers less than 10) and B = (multiples of 3 greater than 3) A:name the communities A and B B:desc

ribe the community A n B​
Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

A n B = {6}

Step-by-step explanation:

Given he sets A = (even numbers less than 10) and B = (multiples of 3 greater than 3)

A = {2, 4, 6, 8}

B = {3, 6, 9, 12, 15...}

A n B includes the numbers are are common to both sets i.e number that we can find in both A and B.

From the set given, we can see that the only number that exists in both set is 6, hence;

A n B = {6}

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98464 rounded to the nearest ten thousand?​
dusya [7]

Answer:

100000

Step-by-step explanation:

4 0
3 years ago
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Simplify: 10x-6(7+x)
Basile [38]

Answer:

4 x − 4 2

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HELP PLS :) <br> What is the slope of the line?
Vedmedyk [2.9K]

Answer:

-1/3

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3 0
3 years ago
2x + 7= -11 please help
Dvinal [7]

Answer:

x = -9

Step-by-step explanation:

You are solving for x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 7 from both sides:

2x + 7 (-7) = -11 (-7)

2x = -11 - 7

2x = -18

Next, divide 2 from both sides:

(2x)/2 = (-18)/2

x = -18/2

x = -9

x = -9 is your answer.

~

6 0
3 years ago
Read 2 more answers
A phone company offers two monthly plans. Plan A costs $10 plus an additional $0.15 for each minute of calls. Plan B costs $30 p
Mkey [24]

First, you need to write to expressions to model each situation:

Plan A: 10+0.15x

Plan B: 30+0.1x


Next, set the expressions equal to each other and solve for x:

10+0.15x=30+0.1x

<em>*Subtract 0.1x from both sides to isolate the variable*</em>

10+0.05x=30

<em>*Subtract 10 from both sides*</em>

0.05x=20

<em>*Divide both sides by 0.05*</em>

x=400


The plans would have the same cost after 400 minutes of calls.


To find how much money the plans cost at 400 minutes, plug 400 into either expression.  We'll use Plan A:

10+0.15(400)

10+60

70


The plans will cost $70.


Hope this helps!

3 0
3 years ago
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