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balu736 [363]
3 years ago
11

I need help on this pleaseee

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
5 0

Answer:

your answer is d, its very similar to c, but its d bc its goes up as a posative, so in that case it will be the (x+6)

Step-by-step explanation:

Hopefully this helped, if not HMU and I will get you a better answer!

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If F= 1.8C + 32, Solve for C if F = 65​
FromTheMoon [43]

Answer:

1.8333

Step-by-step explanation:

F = 1.8C +32

C = (F - 32) ÷ 1.8

= (65-32) ÷ 1.8

=1.833

Just throw all the other things to the right, so that only C is on the left

6 0
3 years ago
Read 2 more answers
Type the missing number in this sequence:<br> C , 38, 28, 18, 8, -2, -12
VMariaS [17]

Answer:

48

Step-by-step explanation:

If you go in the opposite direction, you will notice that each time the numbers go up by ten. So if you counted ten up from 38 it would be 48!

Let me know if this helps! Bya!

8 0
3 years ago
Which expression represents the length, in inches,<br> of the framed picture?
Len [333]

Answer:

No solution is posible from the information provided

Step-by-step explanation:

6 0
3 years ago
. Quantas senhas com 4 algarismos diferentes podemos escrever com os algarismos 1, 2, 3, 4, 5, 6?
ruslelena [56]

Answer:

We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

Step-by-step explanation:

We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.

Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.

Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.

  • Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).
  • Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).
  • Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).
  • Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).

So, the number of passwords with 4 different digits we can write = 6 \times 5 \times 4 \times 3  = 360 possibilities.

Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

4 0
2 years ago
Determine whether the two expressions are equivalent. If so, tell what property is applied.
Hoochie [10]

Answer:

I think its associative property but I am not sure.

Step-by-step explanation:

4 0
3 years ago
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