All categories

2 years ago

Six friends attend a party. They form pairs for a game. How many different pairs are possible?

1 answer:

5 0

Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

10. AEF against BCD

I believe there are 12 combinations... I just can't think of the last 2 though.

You might be interested in

If the height is 24, solve for x. h=-16x^2+8x+48

disa [49]

Answer:

3121

7−

3121

Step-by-step explanation:

h+16^2-8x-48=0

-7x+16x^2-48=0

3121

7−

3121

8 0

2 years ago

Saul wrote the statements shown in the chart.

kupik [55]

<span>The correct answer is option C. i.e. Statements 1 and 2 are postulates because they are true facts.A line must contain atleast two point because then only line can be formed. Similarly, a plane is formed when there are atleast three non-collinear points.</span>

7 0

2 years ago

Fill in Sin, Cos, and tan ratio for angle x. <br> Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .