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MakcuM [25]
3 years ago
9

How do I solve this

Mathematics
1 answer:
Westkost [7]3 years ago
3 0
It should be 55n-34c because there is no other way to keep going.
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Complete the calculation so that each shows the correct sum.
dsp73
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miss. Nikkel want to divide her class of 23 students into 4 equal teams.is this reasonable?why or why not
ella [17]
No, because if you divide 23 / 4, you get 5.75. You can't get 0.75th of a kid. So it is not reasonable, because it is a decimal, and not a whole number.

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6 0
3 years ago
Read 2 more answers
Please help me with this!
Airida [17]

Step-by-step explanation:

→ 42 = 7u - u

→ 42 = 6u

→ 42/6 = u

→ 7 =u or U = 7

<em><u>hope </u></em><em><u>this</u></em><em><u> answer</u></em><em><u> helps</u></em><em><u> you</u></em><em><u> dear</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>take </u></em><em><u>care </u></em><em><u>and</u></em><em><u> may</u></em><em><u> u</u></em><em><u> have</u></em><em><u> a</u></em><em><u> great</u></em><em><u> day</u></em><em><u> ahead</u></em><em><u>!</u></em>

4 0
3 years ago
8640 divided by40 long division
Tju [1.3M]
The answer is 216.
https://photomath.net/s/YLLqQX
7 0
3 years ago
Can a sequence be both arithmetic and geometric?
artcher [175]
An aritmetic sequence is like this
a_n=a_1+d(n-1) where a1=first term and d=common difference

geometric is a_n=a_1(r)^{n-1} where a1=first term and r=common ratio


can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve d(n-1)=(r)^{n-1}
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric
4 0
3 years ago
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