hasAquestion

Well we have right triangles ADC, ABD, ABE, BEC, BHD

triangle ABE is a 30-60-90 and ADC is a 45-45-90

and then im stuck, :( 

some things i did notice were AHE and BHD are 45-45-90 triangles, and since ABE is a 30-60-90 triangle, BAH is 15 degrees,

...

i think i got it

make the altitude for triangle on ABC on segment AB, and since BAC is 60 degrees, HCA would be 30 degrees

If there are 20 marbles in the bag, and the ratio of marbles is 4:1:2:3

adding 4+1+2+3 = 10, which when multiplied by 2 results in 20, which means there are 4 blue marbles.

Im pretty sure the original question is:

How many ordered triples of positive integers \((a,b,c)\) have \(\text{GCF}(a, b, c) = 2020\) and \(\text{LCM}(a, b, c) = 2020^2\)

Because, 2020 has factors of 2, 2, 5, 101 and 4020 has factors of 2, 2, 5, 3, 67 and that wouldn't result in an integer.

but the answer to the "real question" would be to:

Note that the prime factorization of 2020 is \(2^2,5,101\) and that the prime factorization of \(2020^2\) is thus \(2^4 \cdot 5^2 \cdot 101^2\)

Thus the exponents of 2 in the prime factorizations of a,b,c must be 2, m, 4 in some order, where \(2 \le m \le 4\)That way, we can ensure that the GCD has two factors of 2 and the LCM has four factors of 2 If m=2 there are 3 distinct shufflings of (2, 4, 4.) If m=3 there are 6 distinct shufflings of (2,3,4) If m=4 there are 3 distinct shufflings of (2,4,4) So there are 3 allowed arrangements of powers of 2 among a,b ,and c .

The exponents of 5 for a,b,c must be some shuffling of (1, 2, 3) or (1, 2, 2) by the same logic, giving us 6 allowed arrangements of powers of 5. Likewise, there are 6 allowed arrangements for the powers of 101 by similar reasoning.

Thus the total number of ordered triples is the product of these three counts: \(12\times6\times6\) =   432  ordered triples.

https://www.desmos.com/calculator/ecmqlf6gaz

simplifies to

\(x^7+4x^6+8x^5+16x^4+22x^3+19x^2+15x+7\)

and the coefficient will be 15

This is easy

\(\frac{1}{4}+\frac{1}{4*7}+\frac{1}{7*10}\)

is

\(\frac{35}{140}+\frac{5}{140}+\frac{2}{140}\)

which becomes

\(\frac{42}{140}\)

which simplifies to:

\(\frac{3}{10}\)

In class, we derived that

\(\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.\)

Fill in the blanks to make a true equation:

\(\frac{5x}{(x - 1)(x^2 + 2)(x + 7)^3)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2} + \frac{D}{x + 7} + \frac{E}{(x + 7)^2} + \frac{F}{(x + 7)^3}\)

What is the unit conversion from 156.25 to 12.5 inches

https://web2.0calc.com/questions/probability_18785

try it

Easy,

The distance is:

\(\sqrt{3^2+5^2}\)

from length to width

\(\sqrt{\sqrt{3^2+5^2}^2+11^2}\)

from length, to width, to hight

which simplifies to

\(\sqrt{155}\)

\(x^3 + 3x^2 + 7x − 11 = 0\)

simplifies to:

\((x-1)(x^2+4x+11)=0\)

and 

\(x^2+4x+11=0\)

which gives an imaginary number

but for the complex part, 

you get the roots of x as:

\(x=\frac{-4±\sqrt{4^2-4*1*11}}{2}\)

whic becomes

\(x=2±\sqrt{-7}\)

so the roots of 

\(x^3 + 3x^2 + 7x − 11 = 0\)

are

\(x = 1\)

and

\(x=2±\sqrt{-7}\)

I've seen this problum so i'm assuming that the ratio is 3:2, That means if everyone has an equal chance to be president, the ratio of sophomore swimmers to freshman swimmers is 3:2. So 3/(3+2) is 3/5 of the team are sophomores.

The probablility for the first card to be a king is 1/13

the probability for the second card to be hearts is 3/4 likely to be 1/4 or and 1/4 likely to be 3/13

The probablity for the third to be red is 3/4 likely 1/2 or 1/4 likely 6/13

(1/13*3/4*1/4 +1/13*1/4*3/13) *3/4*1/2+

(1/13*3/4*1/4 +1/13*1/4*3/13) *1/4*6/13

=

2601/281216​