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kicyunya [14]
3 years ago
5

 You could get 50-75  points!!!!!!

Mathematics
2 answers:
Masteriza [31]3 years ago
7 0
Any number in between -21 and -20 would suffice in this case. If there are no other conditions, let's claim that Stan's bank account balance is -20.75
romanna [79]3 years ago
6 0
Thanks for the points. We need gracious people like you.

Less than -20 but greater than -21

One possible answer is =  -20.50.

=  -$20.50

That's a possible balance for Stan's bank account account.

The key here is that for negative numbers, the smaller numbers are bigger than the big numbers. 

E.g  -2  is bigger than -3.
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ASAP PLS
VikaD [51]
Man in gonna tell you something but it might be wrong. about 1 or 2 i think?? i'm not good at this
3 0
3 years ago
9 percent of 1000 is....
Fiesta28 [93]
0.9 ..............................
8 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
What is the perimeter ?
natta225 [31]

Step-by-step explanation:

the outer left side is also 50 ft.

in inner (upper) left side is also 13 ft.

the bottom line is 10+20+11 = 41 ft

this is all possible because all the angles are clearly right angles (90°).

so, the perimeter is

2×(10+20+11) + 2×50 + 2×13 = 82 + 100 + 26 = 208 ft

5 0
2 years ago
Which functions would represent the graph?
Murljashka [212]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
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