The larger of the two positive numbers is 8 more than the smaller. The sum of their squares is 104. Find the two numbers.

Mathematics

1 answer:

poizon [28]3 years ago

4 0

Hey there!

Let's set up a system of equations to find our answer. We will use x and y, with y being the larger 1.

y=x+8

x²+y²=104

We plug our value for y into the second equation and we get that y is equal to two. We can then plug this into the first equation and we get 10.

Our two numbers are two and ten.

I hope this helps!

Let's plug our y value into the second equation.

x²+x²+64=104

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There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the

LiRa [457]

Answer:

No. of juniors = 14

No. of seniors = 16

Total students = 30

A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

Since we are asked about arrangement so we will use permutation

Formula : $^nP_r=\frac{n!}{(n-r)!}$

n = 30

r = 5

$^{30}P_5=\frac{30!}{(30-5)!}$

$^{30}P_5=17100720$

So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?

B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

Out of 16 seniors 2 will be selected

So, 3 places are vacant

Remaining students = 30-2 = 28

So, out of 28 students 3 students will be selected

No. of ways = $^{16}P_2 \times ^{28}P_3$

No. of ways = $\frac{16!}{(16-2)!}\times\frac{28!}{(28-3)!}$

= $4717440$

There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.

C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Since we are not asked about arrangement so we will use combination

Out of 16 seniors 2 will be selected

Out of 14 juniors 2 will be selected

Formula : $^nC_r=\frac{n!}{r!(n-r)!}$