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saul85 [17]
2 years ago
10

Please help!! I'm really confused:(

Mathematics
1 answer:
Makovka662 [10]2 years ago
6 0

60\% = 180\\1\%=\frac{180}{60} = 3

100\% - 60\% = 40\%\\40\% = 3*40=120

Answer: 120 cans.

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The first 150 attendees at a
pashok25 [27]
Use proportion or formula or working

Part
____== %_

Whole. 100

Answer: 500
500 people attended the game
3 0
3 years ago
Six students collect buttons for a project. The students have 10, 14, 15, 12, 13, and 14 buttons. They divide the buttons evenly
vovangra [49]

Answer:

13 buttons

Step-by-step explanation:

The first step is to add the buttons received by the 6 students.

= 10 + 14 + 15 + 12 + 13 + 14

= 78 buttons

Therefore the number of buttons received by each students can be calculated as follows

= 78/6

= 13 buttons

Hence each student will receive 13 buttons

6 0
2 years ago
Read 2 more answers
Khan Acad
alexandr402 [8]

Answer:

7.222 (repeating)

Step-by-step explanation:

8 + 10 + 8 + 5 + 4 + 7 + 5 + 10 + 8 = 65

1      2    3    4    5    6    7    8     9

65 / 9 = 7.222 (repeating)

6 0
2 years ago
Mr. Mole's burrow lies 5 55 meters below the ground. He started digging his way deeper into the ground, descending 3 33 meters e
Alexeev081 [22]

Answer:

184,815

Step-by-step explanation:

so my answer my not be right so i would say double check but anyway just multiply 3333 and 555 and this should be your answer.

7 0
3 years ago
For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
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