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Kay [80]
3 years ago
12

3 What is the difference of 21 ? 4

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

The difference is 4–3. If you want the difference between 3 and -4, then the difference is 7.

Step-by-step explanation:

You might be interested in
PLEASE ANSWER -( 1 - 6n )
Slav-nsk [51]

Answer:

6n-1

Step-by-step explanation:

Re-arrange
-(1 - 6n)
-(-6n + 1)

Distrubute
-(-6n -1)
6n - 1

4 0
3 years ago
The diagram shows a polygon. Find the area, in cm², of the whole diagram. ​
Leto [7]

Answer:

Bottom part = 10 * 12 = 120 square centimeters

Top part 8 * 6 = 48 square centimeters

Total Area = 168 square centimeters

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
NEED CORRECT ANSWER...
Misha Larkins [42]

39.50-23.76-2.57-1.49=11.68

take the sales and subtract the expenses

6 0
4 years ago
Add the two expressions -4x + 3 and 2x - 6 enter your answer in the Box PLZ HELP FASt
Fed [463]
-4x+3+2x-6
The answer is -2x-3
3 0
4 years ago
In 4 ​minutes, a conveyor belt moves 800 pounds of recyclable aluminum from the delivery truck to a storage area. A smaller belt
kkurt [141]

Answer:

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

Step-by-step explanation:

Given - In 4 ​minutes, a conveyor belt moves 800 pounds of recyclable

            aluminum from the delivery truck to a storage area. A smaller belt

            moves the same quantity of cans the same distance in 6 minutes.

To find - If both belts are​ used, find how long it takes to move the cans

              to the storage area.

Proof -

Given that,

Larger belt take 4 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Larger belt = 4 min/ shifting

Also,

Smaller belt take 6 minutes to transfer the aluminium from delivery truck to storage area.

⇒ Time taken by Smaller belt = 6 min/ shifting

Now,

As we know that Time is inversely proportional to time, so

Rate of Larger Belt = \frac{1}{4} shifting/min

Rate of Smaller Belt = \frac{1}{6} shifting/min

Now,

Let us assume that , if both belts used then,

Total time taken = x min/shifting

Then, Rate = \frac{1}{x} shifting/min

Now,

\frac{1}{4}  + \frac{1}{6}  = \frac{1}{x}

⇒\frac{6 + 4}{24} = \frac{1}{x}

⇒\frac{10}{24} = \frac{1}{x}

⇒10x = 24

⇒x = \frac{24}{10} = 2.4 minutes

∴ we get

It takes 2 minutes 4 seconds to move the cans to storage area if both the belts are used.

3 0
3 years ago
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