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trapecia [35]
2 years ago
12

Question 2

Mathematics
1 answer:
REY [17]2 years ago
8 0

Answer:

C

Step-by-step explanation:

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A) (-63) ÷ (-9)<br>b) (+72) ÷ (-8)<br><br>​
Juliette [100K]

Answer:

A. +7

B. - 9

Hope this help you

3 0
3 years ago
Please help me with this graph ???
olga_2 [115]
FALSE. It's 5.
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6 0
3 years ago
Read 2 more answers
Which equation is equivalent to 24x = 8x-3?
Elena-2011 [213]

Answer:

uh maybe 16x= -3 or x= -3/16

Step-by-step explanation:

I'm so sorry if this is wrong or not what you're looking for

3 0
4 years ago
Read 2 more answers
One batch of muffins require 1/4 cup of sugar. Kelly wants to make 3/2 batch of muffins. Will she use more than, less than, or a
Alla [95]
She will use more than 1/4 cup of sugar to make 1.5 batches of muffins.  If one batch of muffins requires 1/4 cup of sugar, you can think of the amount of sugar needed being 1/4 cup per batch.  That being said multiply 1.5 batches by 1/4 cup per batch to find cups of sugar.
(1.5 batches)x((1/4 cups)/(1 batch))=0.375 cups of sugar.

You can also think of it as if one batch takes 1/4 cup of sugar, half of a batch would take 1/8 cup of sugar.  Since 1.5 batches is just 1 batch+1/2 batch, you can say the amount of sugar needed is 1/4 cup+1/8 cup which equals 3/8 cup sugar.

I hope this helps you.
5 0
3 years ago
A code on a student ID card begins with 3 letters and ends with 5 digits. How many different ID codes are possible if…the letter
Scorpion4ik [409]

Answer:

The total number of combinations is 2,358,720,000

Step-by-step explanation:

For the letter part of the ID we have 3 letters out of a space of 26 possible letters (a-z), and they can't repeat. For the number part we want to group 5 numbers out of 10 possible algarisms (0-9).So we can make an arrangement for the letters and one for the numbers and multiply them. The arrangment can be done using the following formula:

A(n,k) = (n!)/(n-k)!

Where n is the total number of possibilities and k is the size of the group.

For the letters:

A(26,3) = (26!)/(26-3)! = (26!)/(23!) = (26*25*24*23!)/(23!) = 26*25*24 = 15600

For the numbers:

A(10,5) = (10!)/(10 - 5)! = 10!/5! = (10*9*8*7*6*5!)/(5!) = 10*9*8*7*6*5 = 151200

The total number of combinations is the product of both, so:

combinations = 15600*151200 = 2,358,720,000

4 0
3 years ago
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