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alexandr402 [8]
3 years ago
10

please solve this question

Mathematics
2 answers:
erastovalidia [21]3 years ago
4 0
A c and e are correct I think
____ [38]3 years ago
3 0
A, c, and e are the correct answers
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What is 5.6 equalivent to
Sloan [31]
(this is for fractions!!!!!!!!!!!!)  28/5 or 5 3/5

i think u just make it into 5.6/1 multiply by 10 which gets 56/10 and wala 
but it's not simplified. 56 and 10 is divisible by 2 so u get 28/5 which is actually 5 3/5 (which is 5.6!!)
4 0
4 years ago
"Eight more than twice a number" can be written as what algebraic expression?
Ainat [17]
" 8 more "....means add 8
" twice a number " means multiply the number by 2

ur expression is : 2n + 8
7 0
4 years ago
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Use substitution to solve the system.<br> x=3y+5<br> 5x-4y=14<br> ______________<br> x=<br> y=
nataly862011 [7]
So what you do is go 5(3y+5)-4y=14 which equals 15y+25-4y=14 which turns into 11y+25=14 than you move the 25 over which equals 11y=-11 so y=-1 then you enter -1 in for y which looks like 5x-4(-1)=14 than you solve that which is 5x+4=14 than you move the 4 over and you get 5x=10 than x=2

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8 0
3 years ago
Someone chat w me thru the comments on this question. im grounded and lonely. love you hehe
nikitadnepr [17]

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point yktv

Step-by-step explanation:

3 0
3 years ago
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What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?
joja [24]

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

T_{n}=a+(n-1)d

here a = first term

n = number of term

d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

d = 40/8 = 5

Now T_{23}= a + (n-1)d

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.

4 0
3 years ago
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