All Answers

#1

+1

A bit of a trick question! If f(x) is the constant polynomial 20, then you get |f(0)| = 20, which is the minimum.

GuestJan 6, 2020

#1

+36919

+3

9^(3-x) = 9^2(4-2x)

equate the exponents

3-x = 8-4x

3x = 5

x = 5/3

Another way:

If you take the LOG of both sides you will get this:

(3-x) Log 9 = (4-2x) log 81

(3-x)/(4-2x) = log81/log9

(3-x)/(4-2x) = 2

3-x = 8-4x

-5 = -3x

x = 5/3

ElectricPavlovJan 6, 2020

#1

0

Hint: Find the LCM of 13, 17, and 19

GuestJan 6, 2020

#2

+2862

+2

Why is the answer disliked? What is wrong with it?

CalculatorUserJan 6, 2020

#1

+2

We can solve this problem using the Mobius Inversion Formula. (See https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula)

By the Mobius Inversion Formula,

f(k) = k^3 - (k - 1)^3 + (k - 2)^3 - (k - 3)^3 + ...

= 3(k^2 + (k - 1)^2 + (k - 2)^2 + ...) - 6*(k + (k - 1) + (k - 2) + ...) + 3

= 3*k(k + 1)(2k + 1)/6 - 6k(k + 1)/2 + 3

= k^3 - 3/2*k^2 - 5/2*k + 3.

GuestJan 6, 2020

#1

0

There are 7! = 5040 ways to label the heptagon. There are 4! = 24 ways where the O's are next to an N, so the number of ways where N is not next to an O is 5040 - 24 = 5016.

GuestJan 6, 2020

#3

+36919

+2

Middle number = x the other two numbers are x-1 and x+1

(x-1)^2 + x^2 + (x+1)^2 = 110

x^2 - 2x +1 + x^2 + x^2 + 2x +1 = 110

3x^2 +2 = 110

3x^2 = 108

x^2 = 36

x = 6 x-1 = 5 x+1 = 7 5+6+7 =18

ElectricPavlovJan 5, 2020

#1

+36919

+3

rate x time = distance (or amount burned away)

1 candle / 3hr = 1/3 candle/ hr for rate of the fast burner the other is 1/4 candle/hr

The amount of candle remaining after some time t is 1 candle - rate* t

The slower burner will be twice as long as the faster burner at what point?

1 - 1/4 candle/ hr x t = (1 - 1/3 candle/hr x t) x 2 solve for t

1-1/4 t = 2 - 2/3 t

-3/12 t = 1 - 8/12 t

5/12 t = 1

t = 12/5 = 2.4 hours start 2.4 hours before 4 pm or 1:36 PM

ElectricPavlovJan 5, 2020

#3

+118613

+2

Thanks for the input guest

There are glitches with the newest updates.

Hopefully admin will sort them all out soon.

I can't blame admin for disallowing ad blocking. It is the only way that sites are able to pay for themselves.

MelodyJan 5, 2020

#2

+23247

+1

I have a different way of looking at this problem (that results in a different answer):

Choose two of the 13 different ranks to get the two pairs: ₁₃C₂ = 78

Choose two of the 4 different suits for the first pair: ₄C₂ = 6

Choose two of the 4 different suits for the second pair: ₄C₂ = 6

Choose one of the 44 cards that remain: 44

Multiply these answers together: 78 x 6 x 6 x 44 = 123 552

geno3141Jan 5, 2020

#1

+36919

+2

I think if they were all different it would be 5! ways = 120

but 2 are indistiguishible 'O' s.....so divide by 2 = 60 ways (I think ! )

ElectricPavlovJan 5, 2020

#1

+36919

+2

If you grab 5 randomly (if none match )you COULD get 1 red 1 blue 1 orange 1 black and 1 yellow

this is the worst case scenario Now if you grab ONE more sock...it will HAVE to match ONE of those FIVE

so to GUARANTEE a pair, you will have to grab SIX .

ElectricPavlovJan 5, 2020

#1

+36919

+4

From the first equation:

x^2 -2 y^2 =4 (1)

the second equation tells ya what x^2 is equal to....substitute that value into (1)

4y + 8 -2y^2 = 4

2y^2 -4y - 4 = 0

y^2 -2y -2 = 0 Quadratic formula will tell you y = 1 + sqrt3 and y = 1-sqrt3

Use these values of y ...sub into (1) and solve for x 's......That should get you started....

ElectricPavlovJan 5, 2020

#3

0

No, atlas, LOL the tiles are not 16 feet by 16 feet.

_.

GuestJan 5, 2020

#2

0

Find the sum of three positive consecutive integers a, b, c such that a^2 + b^2 + c^2 = 110.

a, b, & c are consecutive, and a² + b² + c² = 110

A third of 110 is about 33 so start from there.

The closest square is 36, so the next two going up would be 49 and 64.

You can easily see that that total is way too large, try the other direction.

The closest square is 36, so the next two going down would be 25 and 16.

That total is way too small, so try the one below 36 and the one above it.

36 + 25 + 49 = 110 ... Like the Baby Bear's porridge, it's just right.

So a, b, & c are 5, 6, & 7, respectively, and so 5 + 6 + 7 = 18 and there's your final answer.

_.

GuestJan 5, 2020

#1

0

Just try a few numbers such as: 5^2 + 6^2 + 7^2 = 110

GuestJan 5, 2020

#1

-1

Right scalene triangle.

Sides: a = 3.536 b = 3.536 c = 5

Area: T = 6.25
Perimeter: p = 12.071
Semiperimeter: s = 6.036

Angle ∠ A = α = 45° = 0.785 rad
Angle ∠ B = β = 45° = 0.785 rad
Angle ∠ C = γ = 90° = 1.571 rad

Area of the square - area of the triangle =100 - 6.25 =93.75 cm^2 - area of the pentagon ABNMD

GuestJan 5, 2020

#4

0

If your question is about "Power Tower", then 2 and -2 cubed will cancel out, leaving you with 1 + 1 = 2

GuestJan 5, 2020

#1

0

Use the GP formula:
Sum =F x ((1 - R^N) / (1 - R))
6825=F x ((1 - 4^6) / (1 - 4))
6825=F x ((1 - 4096) / (- 3))
6825=F x ((- 4095) / ( - 3))
6825=F x 1365
F =6825 / 1365
F = 5 - First term of the GP

GuestJan 5, 2020

#1

-2

From the formulas

\(\sum_{k = 1}^n k = \frac{n(n + 1)}{2}, \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n+ 1)}{6}\)

it follows that f(k) = k^4/4 + k^3/2 + k^2/4.

GuestJan 5, 2020

#20

+397

+2

I arrived at the now accepted result, but via a different method.

Think of a qualifying sequence as any containing exactly two ones,two twos, and two "others".

Every qualifying sequence has the same probability of occurring, (1/6).(1/6).(1/6).(1/6).(4/6).(4/6)=16/(6^6).

The order doesn't matter.

The number of qualifying sequences will be 6!/(2^3), the three twos to remove the duplication of the 1's, the 2's, and the "others".

The total probability will be the product of these.

That works out to be 5/162, approximately 0.030864.

TiggsyJan 5, 2020

#1

+1

There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton. So: 13*3*12*3*11*4 = 61776 in total.

GuestJan 5, 2020

#3