Ask question

Ask question

All categories

2 years ago

10

I need help on how to do combinations

1 answer:

Veronika [31]2 years ago

3 0

Ok sure! Let's start with the marbles first. We have one red one blue and one green. There is a red and a blue cube. All we do is list the possible combinations of one marble and one cube. The answer would be : 1 blue marble, 1 blue cube. 1 red marble, 1 blue cube. 1 green marble, 1 blue cube. Then again but with the red cube. So in all, there are 6 possible combinations. Hope this helps. ☺️

You might be interested in

Please help quickly <br><br> AB=AD and BC=DC<br> x=?

Arisa [49]

Answer:

x = 110

Step-by-step explanation:

measure of abd is 35, meaning adb is also 35.

so dab + abd + adb is 180

=> x + 35 + 35 = 180

=> x + 70 = 180

=> x = 180-70

=> x = 110

110 degrees is the value of x.

8 0

1 year ago

A taxi charges a fee, plus $1.80 per mile. how far can you go for $12?

EleoNora [17]

12: 1.8= 20/3. => the answer is 20/3 miles

7 0

3 years ago

Hector is dividing students into groups for an HR wants to divide the boys and girls so that each group has the same number of t

Roman55 [17]

Answer:

7 groups, 3 boys and 8 girls

Step-by-step explanation:

Let total number of groups be $x$

and total hikers in one group be $y$

if number of girls and boys in every group is same, then

$\frac{21}{x} + \frac{56}{x} = y$

$\frac{21 + 56}{x} = y\\x*y = 77$ ....(1)

from (1) $x$ will either be 11 or 7

for $x = 11$ the values won't be real.

For $x = 7\\y = \frac{77}{7} \\y = 11$

so there will be 7 groups with 11 hikers in each groups and every group will have $\frac{21}{7} = 3$ boys and $\frac{56}{7} = 8$ girls

6 0

3 years ago

The difference of 48 and a number

scoundrel [369]

48-X or C-48 would be the answer

3 0

2 years ago

I have the question in a screengrab. Thank you, whoever helps me.

nata0808 [166]

To simplify

$\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}$

we need to use the fact that

$\sqrt[4]{x^4}=|x|$

Why the absolute value? It's because $(-x)^4=(-1)^4x^4=x^4$ .

We start by rewriting as

$\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}$

$\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}$

Since $x\neq0$ , we have $\dfrac xx=1$ , and the above reduces to

$\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}$

Then we pull out any 4th powers under the radical, and simplify everything we can:

$\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}$

$\dfrac1{|2y|}\sqrt[4]{3x^2}$

where $y>0$ allows us to write $\dfrac yy=1$ , and this also means that $|y|=y$ . So we end up with

$\dfrac{\sqrt[4]{3x^2}}{2y}$

making the last option the correct answer.

7 0

2 years ago