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a : o : p = 4 : 7 : 9

# of pears sold:           76.50/0.50 = 153

# of oranges sold:       (7*153)/9 = 119

# of apples sold:          (4*153)/9 = 68

I'll let you finish your homework.

\(531n \equiv 1067n \pmod{24}\\ 24*22n+3n \equiv 24*44n+11n \pmod{24}\\ 0+3n \equiv 0+11n \pmod{24}\\ 3n \equiv 11n \pmod{24}\\ 8n \equiv 0 \pmod{24}\\ n=0,3,6,9,etc\\ \text{The smallest positive n is 3}\)

f(x) = x^2 - 3x + 4

f(2x+1) = (2x+1)^2 - 3(2x+1) + 4

now simplify

What is the value of c for which the inequality $x\cdot(3x+1)

Hello Guest!

\(x\cdot(3x+1) \)

What is the inequality called and where is c?

  !

Thank You Melody

I don't know if it's even possible to say that one complex number is greater than another. 

I guess you could say like |3 + 4i| = 5 and |12i + 5| = 13, and 5 < 13. (basically like further away from origin)

=^._.^=

it is 2 - 4x

assuming f(x) is from reals to reals, the range is $[7, +\infty)$

We obtain $\frac{7}{12} \dagger \frac{8}{3} = (7)(8)(\frac{3}{12}) = \boxed{14}$

yeah, inequalities with complex numbers aren't well defined lol

i guess the x >= 0 would imply that

I'm a bit curious, does the x >= 0 imply that x is a real number?

Since something like 3i can't be greater than 0 right?

=^._.^=

spitting facts

i think the problem should state that this holds for all real x

Since x>= 0, we can square this inequality on both sides. 

1 + x >= 2sqrt(x)

x^2 + 2x + 1 >= 4x

x^2 - 2x + 1 >= 0

(x-1)^2 >= 0

A square number always has to be non-negative. 

=^._.^=

(1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 72, 96, 144)==531 - sum of the proper divisors of 288.

y = C1e^−3x + C2e^−2x + C3e^−x

do you mean

y = C_1e^(−3x) + C_2e^(−2x) + C_3e^(−x)

I do not know what you want me to do with this but I could rearrange it as follows.

\(y = C_1e^{−3x} + C_2e^{−2x} + C_3e^{−x}\\~\\ y = e^{-3x}(C_1 + C_2e^{x} + C_3e^{2x})\\~\\ y = \frac{C_1 + C_2e^{x} + C_3e^{2x}}{e^{3x}}\\~\\ \)

What is the sum of the final three digits of the integer representation of $5^{12}$?

Hello Guest!

\(5^{12}=244140\color{blue}625\)

the sum of the final three digits is 13.

  !

The question is wrong

How many integers n between 1 and 10,000 are there such that $7/n$ is a terminating decimal?

I found this easily in google:

"To find out whether a fraction will have a terminating or recurring decimal, look at the prime factors of the denominator when the fraction is in its most simple form. If they are made up of 2s and/or 5s, the decimal will terminate."

So that means that  the desired numbers must be of the form 

 \(2^a*5^b\qquad or \qquad 7*2^a*5^b\)

where a and b are integers greater or equal to 0

I just used excel to help me organize the possibilities.

I also got 78 possibilities

pi / 64

\(\frac{\pi}{4}\div 16 = \frac{\pi}{64}\)

count the  multiples of 8.   (how many are there)

Then put that over 100   (which is the total number numbers.)

Solve |4y + 1| < |2y - 3| - 7

4y+1>=0   when  y>= -1/4   = -0.25

2y-3>=0 when   y>= 1.5

so you have three regions you must solve for

y< -0.25                     -0.25 <= y <  1.5        and       y>=1.25

set up the correct equation (without absolute signs)  for each of those and solve them.

1 will divide any so that is 5 pairs

2 divides 4 and 6 so there is 2 pairs

3 divides 6 so that is 1 pair

so 8 pairs are favourable

there are  6C2 = 15 pairs altogether

so   P(little one divides the larger one) is  8/15

DX = 148/12