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galben [10]
4 years ago
7

(10+43-5)÷6+52 help setup Help me setup

Mathematics
2 answers:
cricket20 [7]4 years ago
6 0

work out brackets furst

10 + 43 - 5= 48

then divide by 6

48 ÷6

then add 52 to the previous answer

52+ 8

=60

lara [203]4 years ago
4 0

Answer:

60

Step-by-step explanation:

(53-5)=48

48÷6+52=

8+52=60

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Solve the division problem using the area model strategy. Upload a picture or video to show your work.
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Answer:

45.333...

Step-by-step explanation:

Try using a long division problem then try the area model strategy

8 0
3 years ago
Kylie is planning to provide online science and Chinese language tutoring sessions. The sciences
Ivahew [28]

Answer:

120 ≤ 20c + 40s

Step-by-step explanation:

(Assuming her name is Kylie, who is giving 40 minute science sessions and 20 minute Chinese sessions.)

The unit for time will be minutes.

Write an equation for time needed for science sessions

40s = t

Write an equation for time needed for Chinese sessions

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Combine the two equations for the total time.

20c + 40s = total time

She does not want the total time to be more than two hours.

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120 ≤ 20c + 40s

8 0
3 years ago
RIGHT TRIANGLES &amp; TRIG!!! <br> Find the value of each variable.
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
Complete the table and show work
Alinara [238K]
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Here, I made a table for it. 

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5 0
3 years ago
Read 2 more answers
The molar mass of AlCl3 is 133.34 g/mol. How many molecules of AlCl3 are there in 2 g?
Mnenie [13.5K]

To solve this problem, we have to apply mole-concept to this and we would need the molar mass of AlCl3.  2g of AlCl3 contains 9.034*10^21 molecules.

<h3>Number of Molecules in 2g of AlCl3</h3>

The molar mass of AlCl3 is given as 133.34g/mol

Data;

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Using the mole concept, we can equate the molar mass to the number of molecules present in the entity.

133.34g = 6.023*10^2^3

If we equate this

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From the calculations above, 2g of AlCl3 contains 9.034*10^21 molecules.

Learn more on mole concept here;

brainly.com/question/21150359

#SPJ1

5 0
2 years ago
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