All Answers 358087 Answers

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100,000 - 999,999 : 112,496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

(1, 1) == 2 , 
(1, 2) == 3 ,
(2, 1) == 3 , 
(1, 3) == 4 , 
(2, 2) == 4 , 
(3, 1) == 4 , 
(1, 4) == 5 , 
(4, 1) == 5 , 
(2, 3) == 5 , 
(3, 2) == 5 , 
(1, 5) == 6 ,
(5, 1) == 6 , 
(2, 4) == 6 , 
(4, 2) == 6 , 
(3, 3) == 6 , 
(1, 6) == 7 ,
(6, 1) == 7 , 
(2, 5) == 7 , 
(5, 2) == 7 , 
(3, 4) == 7 , 
(4, 3) == 7 , 
(2, 6) == 8 , 
(6, 2) == 8 , 
(3, 5) == 8 , 
(5, 3) == 8 , 
(4, 4) == 8 , 
(3, 6) == 9 , 
(6, 3) == 9 , 
(4, 5) == 9 , 
(5, 4) == 9 , 
(4, 6) == 10,
(6, 4) == 10 , 
(5, 5) == 10 , 
(5, 6) == 11 , 
(6, 5) == 11 , 
(6, 6) == 12
(Total = 36)

The roots of the equation are x^3 = 1 are 1, -(1/2) - I root(3)/2  and -(1/2) + I root(3)/2.

x =1 is excluded so x is one of the other two.

It doesn't matter which, since one is the square of the other.

What happens after that ....... .

could very well might be a chatgpt-generated solution so wanted to see other people's opinions on this

I'd like to see this one solved too (with plenty of explanation).   

Anyone interested?   

Right triangle XYZ has legs of length XY = 12 and YZ = 6.  Point D is chosen at random within the triangle XYZ.  What is the probability that the area of triangle XYD is at most 20?

If point D is anywhere in the region    XYAC  then the area of  XYD will be at most 20.   Can you work out why?

So the prob that XYD<=20    is          Area of XYAD divided by  Area of  XYZ

Try proofreading your question.

I expect you can work out (or count)  part a  just as well as any of us can.  

There are 9 individual digits then 2 digits for each number from 10 to 99.     How many is that?

Part b.  Count how many zeros there are and put that number of the total number of digits.

Part c.  Do the other 2 parts then look for patters that you can add together.

To solve this problem, we can use the concept of permutations with repetition. We have two groups of beads: one group with two identical beads and another group with four identical beads. Let's represent the two groups as "A" and "B", respectively. Then, we can use the following formula:

(a) To describe the graph of the parametric equations, we can eliminate the parameter t to obtain an equation in terms of x and y. From the given equations, we have:

sin pit = x - 1 + cos pit
cos pit = (y - 3sin pi*t)/2

Substituting the first equation into the second equation, we get:

cos pit = (y - 3(x - 1 + cos pit))/2
2cos pit = y - 3x + 3
cos pit = (y - 3x + 3)/2

Substituting this into the first equation, we get:

sin pit = x - 1 + (y - 3x + 3)/2
2sin pit = 2x - 2 + y - 3x + 3
sin pit = x + y - 1

Therefore, the graph of the parametric equations is the set of all points (x, y) that satisfy the equation x + y - 1 = sin pi*t, where t ranges over all real numbers.

(b) To describe the motion of the particle as t ranges from 0 to 2, we can evaluate x and y at t = 0 and t = 2, and plot the resulting points.

At t = 0, we have:

x = 1 + sin 0 - cos 0 = 1
y = 3sin 0 + 2cos 0 = 2

At t = 2, we have:

x = 1 + sin pi2 - cos pi2 = 1
y = 3sin pi2 + 2cos pi2 = -1

Therefore, the particle moves horizontally along the line y = 2, then returns to its starting point by moving horizontally along the line y = -1.

(c) To find a parametrization that matches the graph of part (b) but with a different motion, we can use a different function for x or y. For example, we can use:

x = 1 + sin pit
y = 2cos pi*t

You got it :)

The coordinate plane contains points A, B, and C. The constant k then changes to 32.

Let P be (x,y)

= Pa^2+Pb^2+pc^2

= (x-4)^2+(y+1)^2+(x-6)^2+(y-2)^2+(x+1)^2+(y-2)^2

= 3x^2+3y^2-18x-6y+64

= 3(x^2+y^2-6x-2y)+62

= 3((x-3)^2+(y-1)^2)+32

This shows that if Q = (3,1), Pq^2=(x-3)^2+(y-1)^2

And so pa^2+pb^2+pc^2=3pq^2+32

k = 32

Start by counting all the perfect powers of non perfect powers of 2, 3, 5, 6, 7, 10 that are below 100. How many perfect powers we have, how many numbers we need to add onto 100. Then we repeat this process with the numbers from 100 to the new number we get until we finally have no numbers left (it wont be so confusing later on)

First we look at multiples of 2. There is 2^0 (we will only be counting 1 once) 2^2, 2^3, all the way to 2^6 ( 64). There is 6 numbers

Then look at multiples of 3. There is  3^2, 3^3, 3^4 There is 3 numbers

we skip 4 because 2 already includes all of these

From here, 5, 6, 7, and 10 all only have 1 multiple. 

We find the total number of multiples under 100 by doing 6 + 3 + 4  = 13 

Adding 13 to 100 we get 113.

We find that there are no numbers between 100 and 113 that are perfect powers therefore our answer is, 113

To solve for the minimum value of the magnitude of a+bx+cx^2 where a,b, and c are distinct integers and x^3=1 and x≠1, we can use the fact that x^3=1 to write x^3-1=0, which can be factored as (x-1)(x^2+x+1)=0. Since x is not equal to 1, then the equation x^2+x+1=0 must hold true.

Next, we can use the fact that x is a complex number to write x in terms of its real and imaginary parts, as x=r(cosθ+isinθ), where r is the magnitude of x and θ is the angle that x makes with the positive real axis. Substituting x into the equation x^2+x+1=0 and using Euler's formula, we can write the equation as:

r^2(cos2θ + cosθ) + r(sin2θ + sinθ) + 1 = 0

This is a quadratic equation in r, so we can solve for r using the quadratic formula:

r = [-b ± sqrt(b^2 - 4ac)] / 2a

where a = cos2θ + cosθ, b = sin2θ + sinθ, and c = 1. Note that a,b, and c are all real numbers.

Since we want to find the minimum value of the magnitude of a+bx+cx^2, we can write this expression in terms of r and θ and then minimize it with respect to θ. Using the formula for x and simplifying the expression, we get:

|a+bx+cx^2| = |a + brcosθ + crcos2θ + brsinθ + crsin2θ|

We can then use the fact that sin2θ = 2sinθcosθ and cos2θ = cos^2θ - sin^2θ to rewrite this expression as:

|a+bx+cx^2| = |(a + bcosθ) + (bsinθ + ccosθ)r + csin^2θ|

This is minimized when θ=pi/4, to give a minimum value of 2*sqrt(2).

ya the answer is 2

Do not cheat on A O P S homework.

You're on the right track! From the equation z^n = 1, we know that z must be an nth root of unity, so we can write:

z = e^(2πik/n)

where k is an integer such that 0 ≤ k < n. This gives us n distinct solutions for z, one for each value of k.

To find the solutions for (z+1)^n = 1, we can first notice that if we define w = z+1, then the equation becomes w^n = 1. So we know that:

w = e^(2πik/n)

for some integer k between 0 and n-1. Substituting z+1 for w, we get:

z + 1 = e^(2πik/n)

which we can rearrange to get:

z = e^(2πik/n) - 1

Now we have n distinct solutions for z, one for each value of k between 0 and n-1.

To write these solutions in exponential form, we can use Euler's formula:

e^(iθ) = cos(θ) + i sin(θ)

So we can write each solution as:

z = e^(2πik/n) - 1 = cos(2πk/n) + i sin(2πk/n) - 1

= (cos(2πk/n) - 1) + i sin(2πk/n)

= -2 sin²(πk/n) + i 2 sin(πk/n) cos(πk/n)

where we've used the identity cos(θ) - 1 = -2 sin²(θ/2) and the double-angle formula sin(2θ) = 2 sin(θ) cos(θ).

So the n solutions for z in exponential form are:

z = -2 sin²(πk/n) + i 2 sin(πk/n) cos(πk/n)

for k = 0, 1, ..., n-1.

The only functions that work are f(x) = x and f(x) = -x.  So there are two functions that are their own inverses.

The value of b is 8.

Since -5 and 6 are roots of the equation, we know that (x+5) and (x-6) are factors of the cubic polynomial. This can be seen by using the factor theorem:

f(-5) = a(-5)^3 + b(-5)^2 + c(-5) + d = 0 f(6) = a(6)^3 + b(6)^2 + c(6) + d = 0

Since both expressions are equal to zero, we can use them to write:

a(-5)^3 + b(-5)^2 + c(-5) + d = a(6)^3 + b(6)^2 + c(6) + d

Simplifying this equation, we get:

-125a + 25b - 5c = 216a + 36b + 6c

Collecting like terms and simplifying further, we get:

341a + 61b = 11c

Now, using the fact that (x+5) and (x-6) are factors of the cubic polynomial, we can write:

f(x) = a(x+5)(x-6)(x-r)

where r is the third root of the polynomial.

Expanding the right-hand side of this expression, we get:

f(x) = a(x^2-x-30)(x-r)

Using the fact that the sum of the roots of a cubic polynomial is given by -b/a, we have:

-5 + 6 + r = -b/a

Simplifying this equation, we get:

r = -1 + b/a

Now, we can use the equation we derived earlier, 341a + 61b = 11c, to eliminate b and express r in terms of a and c:

r = -1 + b/a = -1 + (11c - 341a)/(61a) = (-342a + 11c)/(61a)

Therefore, the expression for the third root in terms of only a and c is:

r = (-342a + 11c)/(61a)