All Answers 359395 Answers

Let the common difference be d, then

\(\displaystyle a_{1}=10, \\ a_{2} = 10 + d, \\ a_{3}=10 + 2d, \\ \dots \\ a_{n}=10 + (n-1)d, \\ \text{etc.} \)

\(\displaystyle a_{a_{2}}=a_{10+d}=10 +(9+d)d=d^{2}+9d+10 = 100, \\ d^{2}+9d-90=0,\\ (d-6)(d+15)=0,\\ d=6, (\text{or } d=-15). \)

So then, for example, and in anticipation,

\(\displaystyle a_{3} = 10 + (3-1).6 = 22, \\ a_{22}=10+(22-1).6= 136.\)

Finally,

\(\displaystyle a_{a_{a_3}}=a_{a_{22}}=a_{136}=10+135.6=10+810 = 820.\)

I saw that one but the answer is estimated. Can you give an exact answer please? Thanks :D

1391/0.07+1

1300 ans

A 101-gon is just like an octagon but with 101 sides

Since no three vertices of an octagon are colinear, then any possible triplet forms a triangle. So the number of non-congruent triangles is equal to all possible ways to choose three vertices of an octagon, where rotations and reflections still count as the same object.

One simple way to count them is to realize that if you have three vertices, there are three gaps between them. Each of these three gaps contains 0 to 5 of the remaining vertices and all three together have 5 vertices.

And since there are only three of them, their order is irrelevant (it would be different if there were four of them).

So, the possibilities are:

(0,0,5)
(0,1,4)
(0,2,3)
(1,1,3)
(1,2,2)

There are $5$ non-congruent triangles that can be formed from vertices of a regular octagon.

Ohhhh, I guess I was over thinking it. Thank you!! 

nevermind, i got it!

Answer #3, #2 pretty much gets you there.

There are 20 possibilities for the final two digits and that will fix the first two digits as well.

For the middle three digits, there are 10 possibilities for the middle digit and 10 for the (identical) digit either side of that, so 10*10=100 possibles.

Put those together and you have 20*100 = 2000 possible 7 digit palindromes, (as was stated in #1). 

Set y = 0 and solve the resulting equation for t.  There will be two values; choose the only sensible one.

y=-6t^2+43t

In how many seconds will the projectile first reach 77 feet in height?

Hello Guest!

\(77=6t^2+43t\\ -6t^2+43t-77=0\ |\ /(-6)\\ t^2-\frac{43}{6}+\frac{77}{6}=0\\ t=\frac{43}{12}\pm \sqrt{(\frac{43}{12})^2-\frac{77}{6}}\\ \color{blue}t\in \{3.5,3.\overline 6\}\)

In 3.5 seconds will the projectile first reach 77 feet in height.

  !

There are 5 cases: 

4 - 0 - 0 - 0 

3 - 1 - 0 - 0 

2 - 2 - 0 - 0

2 - 1 - 1 - 0

1 - 1 - 1 - 1

For the first case, there is only 1 way. 

In the second case, there are \({4 \choose 3} = 4 \) ways.

In the third case, there are \({4 \choose 2} = 6\) ways.  

In the fourth case, there are \({4 \choose 2} \times 2 = 12\) ways. 

For the fifth case, there is only 1 way.

So, there are \(1 + 4 + 6 + 12 + 1 = \color{brown}\boxed{24}\) ways. 

Simplify: \((qp + 4p + 3q + 12) - (qp + 2p + q + 2) = 44 \)

Combine like terms: \(2p + 2q + 10 = 44\)

Subtract 10 from both sides: \(2p + 2q = 34\)

Divide both sides by 2: \(q + p = \color{brown}\boxed{17}\)

I know the divisibilty rules. i just don't understand how to include the palindrome part of the question

possible last 2 digits(divisibility rule for four):

00

04

08

12

16

20

24

28

32

36

40

44

48

52

56

60

64

68

72

76

80

84

88

92

96

25 possibilities. Out of these, the ones ending with 0 don't work. (Can you see why?)

So, 00, 20, 40, 60, and 80 don't work. 

20 possibilities.

factor it.

(3y + 8)(y - 3)

a = 8

b = -3

a - b = 11

Ohhh thank you! I understand how to solve now!

The nearest hundred would be 0

although i think you might be looking for the nearest hundreth. Either way 8.55 to the nearest hundreth is 8.55.
the thousanth place value for 8.55 is 0 so the hundreth place would stay the same.

Thanks.  You're almost right!  I added the square root of 75 instead of subtracting it.  I've now edited and corrected.

I think you added the square root of 12 instead of subtracting

Like so:

\(\sqrt{12}-\sqrt{75}+\sqrt{147}=\sqrt{3*4}-\sqrt{3*25}+\sqrt{3*49}=(2-5+7)*\sqrt{3}=4\sqrt{3}\)

Other than a computer finding it how could I approch it?

Go to this link and see the formula used (Formula(9)) under "distinct"   and   "identical" :

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php