All categories

3 years ago

Help please please please

Mathematics

2 answers:

blagie [28]3 years ago

8 0

Answer:

1/3

HELP ME IN THIS QUESTION:

In art class, the students each made rectangular quilts using 24 paper squares. They then glued a button onto each square and decorated the squares with markers.

1. Create all of the possible arrays for modeling a rectangular quilt with the 24 paper squares. Write an equation to accompany each array. You can use the attached pre-made quilt squares by cutting and pasting them, or you can draw the arrays with your own designs on a separate sheet of paper

ONLY REAL QUESTIONS

galben [10]3 years ago

5 0

Answer: 1/3

Step-by-step explanation: The slope goes up 1 and over 3 each time. Rise over run : 1/3

You might be interested in

-2b+2(b-10)=2(10+5b)

alina1380 [7]

Answer:

b= -4

Step-by-step explanation:

On both sides of the equation has distributive property. We need to solve that first. Let's start with the left side of the equation.

*Positive multiplied by a negative equals a negative

-2b+2(b-10)= 2(10+5b)

-2b+2b-20= 2(10+5b)

-20= 2(10+5b)

Now let's solve the right side of the equation

-20= 20+10b

-20 -20

-40 = 10b

-40/10 = 10b/10

b= -4

Hope this helps!

5 0

3 years ago

How many ways can 100 be written as the sum of distinct integers from the set {1, 2,..., 15}?

konstantin123 [22]

Answer:

48 ways

Step-by-step explanation:

Let me take a guess

S₁_₁₅ = (1+15)*7 + 8 = 120

There are 48 combinations of distinct digits from 1 to 15 to make 20

120-20=100

So every 20 has a corresponding 100

I wish I got it right, otherwise report it.

4 0

2 years ago

Change 101/22 into a mixed fraction

ss7ja [257]

4 13/22

hope this helps :)

3 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,

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