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Kay [80]
2 years ago
9

PLEASE HELPPPP THIS ISNON MY QUIZ

Mathematics
1 answer:
Bess [88]2 years ago
6 0
The answer is d. i hope this helps!
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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
600 millimeters of rain fell in 30 minutes what is the unit rate for millimeters per minute
sleet_krkn [62]
600 mm of rain / 30 min = 50 mm of rain per minute
3 0
3 years ago
Solve for m. m=-(4+m)+2
pishuonlain [190]

Answer:

m= -1

Step-by-step explanation:

Order summands

2

Apply negation to the value inside parentheses

3

Add the numbers

4

Add 

mm

 to both sides of the equation

5

Simplify

3 0
2 years ago
Read 2 more answers
What is the missing number?38/152 19/?
yulyashka [42]

Answer:

76

Step-by-step explanation:

Hope this helps

8 0
2 years ago
Which values of c will cause the quadratic equation –x2 3x c = 0 to have no real number solutions? check all that apply.
nika2105 [10]

The given quadratic equation will not have any real solution for c<-9/4.

The given quadratic equation is:

-x^{2} +3x+c=0

<h3>What is a quadratic equation?</h3>

Any equation of the form ax^{2} +bx+c=0 is called a quadratic equation with a≠0.

In order to have no real solution, the discriminant of a quadratic equation will be less than zero.

D < 0

3^{2} -4(-1)(c) < 0

9+4c < 0

c < -\frac{9}{4}

For c < -\frac{9}{4} the given quadratic equation will have no real solutions.

Hence, the given quadratic equation will not have any real solution for c<-9/4.

To get more about quadratic equations visit:

brainly.com/question/1214333

5 0
1 year ago
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