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tekilochka [14]
3 years ago
10

Can someone help me with these math problems

Mathematics
2 answers:
LiRa [457]3 years ago
8 0
The first one is 40 pie
WITCHER [35]3 years ago
4 0
The second one is 1,456 pie
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Expand and simply <br>a) 2(4c+5d)+3(c-3d) <br>b) 6(3m+n)-4(m-n)<br>​
dezoksy [38]

Answer:

a. 11x + d

b. 15m + 10n

Step-by-step explanation:

a. 2(4c+5d) + 3(x-3d)

Break the brackets:

2*4c + 2*5d + 3*x - 3*3d

= 8x + 10d + 3x - 9d

Collect like termsL

= 8x + 3x + 10d - 9d

= 11x + 1d

b. 6(3m+n) - 4(m-n)

Break the brackets:

= 6*3m + 6*n - 3*m - 4*(-n)

= 18m + 6n - 3m - (-4n)

= 18m -3m + 6n + 4n

= 15m + 10n

Hope this helped :3

3 0
3 years ago
Please guys help me I need help to get an A+
Sergeeva-Olga [200]
The “math sentence”: To find the answer, I need to divide the six chocolate bars by 3/4.

To model the situation, you need to draw 6 squares divided into quarters. Color in 3 of the quarters in each square.

Finally, write this for the solution sentence:
By giving 3/4 of a chocolate bar to each of her friends, Mrs. Lopes will be able to share her 6 chocolate bars with 8 different friends.


Hope you get that A+!
3 0
3 years ago
It cost $13.50 to but 3 used video games how much will it cost to buy 1
fomenos

Answer:

4.50

Step-by-step explanation:

thats really easy bro just divide $13.50 by 3 and you get your answer

7 0
2 years ago
Read 2 more answers
If you place a 33-foot ladder against the top of a building and the bottom of the ladder is 26 feet from the bottom of the build
Firdavs [7]

Answer:

20

I hope that helped! :)

5 0
2 years ago
The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacen
kompoz [17]

Answer:

\fbox{\begin{minipage}{4em}36 ways\end{minipage}}

Step-by-step explanation:

<em>Step 1: Re-state the problem in an easier way to set up a permutation problem</em>

"The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacent in the circle is a multiple of seven"

The series above can be represented again, in form of:

7a + 1, 7b - 1, 7c + 1, 7d - 1, 7e + 1, 7f - 1

with a, b, c, d, e, f are non-negative integers.

=> 3 numbers are multiply of 7 plus 1

=> 3 numbers are multiple of 7 minus 1

<em>Step 2: Perform the counting:</em>

For the 1st number, there are 6 ways to select

To satisfy that each pair of numbers creates a multiple of 7, then:

For the 2nd number, there are 3 ways left to select

For the 3rd number, there are 2 ways left to select

For the 4rd number, there are 2 ways left to select

For the 5th number, there are 1 way left to select

For the 6th number, there are 1 way left to select

=> In total, the number of possible ways to select:

N = 6 x 3 x 2 x 2 x 1 x 1 = 72

However, these numbers are located around a circle, each option is counted twice.

=> The final number of possible ways:

N = 72/2 = 36

Hope this helps!

:)

3 0
3 years ago
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