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But, doesn't he/she say "The sum of first n terms" =n^2 + 5n, which implies "several terms" summing up to n^2 + 5*n ?? Otherwise, he/she could have said " The sum of the FIRST term" =n^2 + 5n.

ok :/

Similar to the ship question above in the forum listing:      d  =   50 t     

the ship travels 25 miles (d)   each hour   (t)  

d = 25 t     or     t = d/25            Cannot really give you a hint.....so there is the answer 

cylinder A has a volume of \(\left(\frac{d}{2}\right)^2\pi h = 14^2\pi \cdot 15 = 2940\pi\) and cylinder B has a volume of \(\left(\frac{d}{2}\right)^2\pi h = 11^2\pi \cdot 159= 2299\pi\). this means that cylinder B is \(\frac{2299}{2940} \cdot 100 \approx 0.7819 \cdot 100 \approx 78.2\%\) percent of cylinder A, which means cylinder A will end up being 

\(100-78.2=\boxed{21.8\%}\)

percent full after the pumping.

https://web2.0calc.com/questions/your-dog-weights-5-times-as-much-as-your-cat-your

The vertical distance 300ft and the horizontal distance are the legs of a right traingle

the tangent of the 15^o  =  opp/adj = 300/horiz     horiz is what you are looking for

tan 15 = 300/horiz       solve for horiz

for a number to be divisible by 25, it must end in either 00 (making it a multiple of 100), 25 (a multiple of 100 + 25), 50 (a multiple of 50), or 75 (a multiple of 100 + 75). since 0 isn't in our digits, the digits must be in the form

\(\times\times\times\times25\)

or

\(\times\times\times\times75\)

where \(\times\) represents any digit not used. there are \(4!\) ways to arrange the digits that fit the first form and \(4!\) ways for the second, so we have a total of 

\(4! + 4! = 24 + 24 =\boxed{48}\)

ways to arrange the digits.

But how far do you want to go? up to 1000, or 1000000 or 1000000000........etc. You have to have a limit !.

The sum of the first term is just the first term. So if we plug in 1 to the equation, we'll find the first term which is (1^2) + 5 * 1 = 6. From this I can already see the difference in our solutions.

The sequence is:
3, 6, 9, 12, 15, 18, 21 =n^2 + 5*n=7^2 + 35=84
Therefore the 1000th term =First term + 3*(1000 - 1)
                                                  =3 + (3*999)
                                                   =3 + 2,997
                                                   =3,000 - which is 1000th term.

Sorry but both answers are wrong! :(

Find slope    = -2

y = -2x + 13   is the equation of the line

2x + y = 13     divide by 13

x / (13/2)  + y/13 = 1

a =4    and     b=396

The list of numbers is just even numbers starting at 6, so the 1000th term would be 2006.

Nah I needed to use a cool symbol so I found the spacer online. :D

Probably looks something like this

d = 5 c

d = c + 20       solve.....

I count 89 such numbers as follows:

GCD of 1 and 100 = 1
GCD of 2 and 100 = 2
GCD of 3 and 100 = 1
GCD of 4 and 100 = 4
GCD of 5 and 100 = 5
GCD of 6 and 100 = 2
GCD of 7 and 100 = 1
GCD of 8 and 100 = 4
GCD of 9 and 100 = 1
GCD of 11 and 100 = 1
GCD of 12 and 100 = 4
GCD of 13 and 100 = 1
GCD of 14 and 100 = 2
GCD of 15 and 100 = 5
GCD of 16 and 100 = 4
GCD of 17 and 100 = 1
GCD of 18 and 100 = 2
GCD of 19 and 100 = 1
GCD of 21 and 100 = 1
GCD of 22 and 100 = 2
GCD of 23 and 100 = 1
GCD of 24 and 100 = 4
GCD of 26 and 100 = 2
GCD of 27 and 100 = 1
GCD of 28 and 100 = 4
GCD of 29 and 100 = 1
GCD of 31 and 100 = 1
GCD of 32 and 100 = 4
GCD of 33 and 100 = 1
GCD of 34 and 100 = 2
GCD of 35 and 100 = 5
GCD of 36 and 100 = 4
GCD of 37 and 100 = 1
GCD of 38 and 100 = 2
GCD of 39 and 100 = 1
GCD of 41 and 100 = 1
GCD of 42 and 100 = 2
GCD of 43 and 100 = 1
GCD of 44 and 100 = 4
GCD of 45 and 100 = 5
GCD of 46 and 100 = 2
GCD of 47 and 100 = 1
GCD of 48 and 100 = 4
GCD of 49 and 100 = 1
GCD of 51 and 100 = 1
GCD of 52 and 100 = 4
GCD of 53 and 100 = 1
GCD of 54 and 100 = 2
GCD of 55 and 100 = 5
GCD of 56 and 100 = 4
GCD of 57 and 100 = 1
GCD of 58 and 100 = 2
GCD of 59 and 100 = 1
GCD of 61 and 100 = 1
GCD of 62 and 100 = 2
GCD of 63 and 100 = 1
GCD of 64 and 100 = 4
GCD of 65 and 100 = 5
GCD of 66 and 100 = 2
GCD of 67 and 100 = 1
GCD of 68 and 100 = 4
GCD of 69 and 100 = 1
GCD of 71 and 100 = 1
GCD of 72 and 100 = 4
GCD of 73 and 100 = 1
GCD of 74 and 100 = 2
GCD of 76 and 100 = 4
GCD of 77 and 100 = 1
GCD of 78 and 100 = 2
GCD of 79 and 100 = 1
GCD of 81 and 100 = 1
GCD of 82 and 100 = 2
GCD of 83 and 100 = 1
GCD of 84 and 100 = 4
GCD of 85 and 100 = 5
GCD of 86 and 100 = 2
GCD of 87 and 100 = 1
GCD of 88 and 100 = 4
GCD of 89 and 100 = 1
GCD of 91 and 100 = 1
GCD of 92 and 100 = 4
GCD of 93 and 100 = 1
GCD of 94 and 100 = 2
GCD of 95 and 100 = 5
GCD of 96 and 100 = 4
GCD of 97 and 100 = 1
GCD of 98 and 100 = 2
GCD of 99 and 100 = 1

a + 15 r = $ 70

a + 12 r = 61        Multiply this equatin by -1     -a - 12 r = -61    add it to the first equation

3 r = 9

r = $ 3       now you can use this value of 'r' to plug into either of the equations to find 'a'   admission cost

If they are parallel, they will have the same slope, m

   arrange the equations in the form   y = mx+b   to see if the 'm' s   are equal

    one  is in this form already    m = 5

2y-10x+3 = 0

2y= 10x-3

y = 5x - 3/2       Yep m= 5 on this one too......