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mote1985 [20]
3 years ago
12

The can is 12cm high and diameter is 8cm how much does it hold

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
5 0
I cannot answer this question because I don't know what liquid or item the can is holding or the formula
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A skier is trying to decide whether or not to buy a season ski pass. A daily pass costs ​$77. A season ski pass costs ​$450. The
n200080 [17]

Answer:

5

Step-by-step explanation:

Given:

  • Daily pass = $77
  • Season ski pass = $450
  • Skis = 20

77 + 20 = 97

450 + 20 = 470

Keep multiplying 97 with random numbers until it passes 470.

97 × 4 = 388

97 × 5 = 485

So, the skier will have to pay daily passes 5 times to be more expensive than the season pass.

Hope this helped.

3 0
2 years ago
Imma need sum help please
MrRa [10]

Answer:

Step 5

Step-by-step explanation:

The mistake is in step 5.

The previous step was

step \: 4 : 2 + 10 \div 2

The order operations, PEDMAS, must be applied:

We have Addition and Division here,

Using PEDMAS, we must divide first to get:

step \: 5 : 2 + 5

We can now add to get:

step \: 6: 7

Therefore the mistake occurred at step 5

8 0
3 years ago
S = ut + 12 at?<br> u= 10<br> a = -2<br> t =<br> Work out the value of s.
Stella [2.4K]
S = (10 x1 /2) + (1/2x- 2x 1/2²)
5 + (-0.25)
= 4.75
7 0
3 years ago
Identify the value of the 4 in the number : 45,348
AnnyKZ [126]
It is 4000 that is the correct answer
4 0
3 years ago
Read 2 more answers
If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930
Free_Kalibri [48]
Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
s=ut+ \frac{1}{2} at^2

But, we assume that the train accelerates from rest, thus
s=\frac{1}{2} at^2 \\  \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\  \\ \Rightarrow t^2=534.48 \\  \\  \Rightarrow t=\sqrt{534.48}=23.12

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.
6 0
3 years ago
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