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lara [203]
2 years ago
15

How many ways are there to distribute the cookies if there are six choco- late cookies, six oatmeal cookies and six red velvet c

ookies
Mathematics
1 answer:
tangare [24]2 years ago
7 0

Answer:

6

Step-by-step explanation:

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Anyone ?? will mark as brainliest
Komok [63]

5+5(x+4)\leq20\\\\
5(x+4)\leq15\\\\
x+4\leq3\\\\
x\leq-1

6 0
3 years ago
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Which equation has a constant of proportionality equal to 2?
Blababa [14]

Answer:

A) y = 10/5x

10 divided by 5 = 2

Step-by-step explanation:

I hope it helps ❤❤

3 0
3 years ago
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Rebecca and dan are biking in a national park for three days they rode 5 3/4 hours the first day and 6 4/5 hours the second day
likoan [24]

Answer:

Rebecca and Dan need to ride 7\frac{9}{20}\ hrs. on the third day in order to achieve goal of biking.

Step-by-step explanation:

Given:

Goal of Total number of hours of biking in park =20 hours.

Number of hours rode on first day = 5\frac34 \ hrs.

So we will convert mixed fraction into Improper fraction.

Now we can say that;

To Convert mixed fraction into Improper fraction multiply the whole number part by the fraction's denominator and then add that to the numerator,then write the result on top of the denominator.

5\frac34 \ hrs. can be Rewritten as \frac{23}{4}\ hrs

Number of hours rode on first day = \frac{23}{4}\ hrs

Also Given:

Number of hours rode on second day = 6\frac45 \ hrs

6\frac45 \ hrs can be Rewritten as \frac{34}{5}\ hrs.

Number of hours rode on second day = \frac{34}{5}\ hrs.

We need to find Number of hours she need to ride on third day in order to achieve the goal.

Solution:

Now we can say that;

Number of hours she need to ride on third day can be calculated by subtracting Number of hours rode on first day and Number of hours rode on second day from the Goal of Total number of hours of biking in park.

framing in equation form we get;

Number of hours she need to ride on third day = 20-\frac{23}{4}-\frac{34}{5}

Now we will use LCM to make the denominators common we get;

Number of hours she need to ride on third day = \frac{20\times20}{20}-\frac{23\times5}{4\times5}-\frac{34\times4}{5\times4}= \frac{400}{20}-\frac{115}{20}-\frac{136}{20}

Now denominators are common so we will solve the numerator we get;

Number of hours she need to ride on third day =\frac{400-115-136}{20}=\frac{149}{20}\ hrs \ \ Or \ \ 7\frac{9}{20}\ hrs.

Hence Rebecca and Dan need to ride 7\frac{9}{20}\ hrs. on the third day in order to achieve goal of biking.

8 0
3 years ago
Suppose that the functions q and r are defined as follows.
satela [25.4K]

Answer:

(r o g)(2) = 4

(q o r)(2) = 14

Step-by-step explanation:

Given

g(x) = x^2 + 5

r(x) = \sqrt{x + 7}

Solving (a): (r o q)(2)

In function:

(r o g)(x) = r(g(x))

So, first we calculate g(2)

g(x) = x^2 + 5

g(2) = 2^2 + 5

g(2) = 4 + 5

g(2) = 9

Next, we calculate r(g(2))

Substitute 9 for g(2)in r(g(2))

r(q(2)) = r(9)

This gives:

r(x) = \sqrt{x + 7}

r(9) = \sqrt{9 +7{

r(9) = \sqrt{16}{

r(9) = 4

Hence:

(r o g)(2) = 4

Solving (b): (q o r)(2)

So, first we calculate r(2)

r(x) = \sqrt{x + 7}

r(2) = \sqrt{2 + 7}

r(2) = \sqrt{9}

r(2) = 3

Next, we calculate g(r(2))

Substitute 3 for r(2)in g(r(2))

g(r(2)) = g(3)

g(x) = x^2 + 5

g(3) = 3^2 + 5

g(3) = 9 + 5

g(3) = 14

Hence:

(q o r)(2) = 14

8 0
2 years ago
If you're room is 86 inches and you're bed is 60 inches can you put a 80 centimeters chest fit next to the bed
Neko [114]

Answer:

yes i think so

Step-by-step explanation:

Queen 60" X 80" 152 X 203.5cm

3 0
2 years ago
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