All Answers 356067 Answers

A question similar to this has been answered:

https://web2.0calc.com/questions/help-please_8381

If you can write the latex, or tell me what numbers are in the problem, I can help you if needed.

Can you please write who many days... I don't think the latex is showing....

\(8\cdot \sqrt{20^2+r^2}=\left(r-20\right)\left(r+20\right)\)

\(64\cdot 400+49r^2=r^4-800r^2+400^2=0\)

\(r^2= \left(\sqrt{\frac{849+\sqrt{183201}}{2}}\right)^2\)

\(r^2= approx.638.5, r = approx.25.3\)

1/2w+9                

\(12*11*10=1320\)

This question has already been answered on StackExchange. It's a pretty neat solution imo. 

Link: https://math.stackexchange.com/questions/1463896/for-1-leq-n-leq-100-how-many-integers-are-there-such-that-fracnn1-is-a

For cos B change the formula so that A → B, a → b, b → c, c → a

See if you can then work out what to do for cos C.

Hello this question was answered before: 

Here is the link:

https://web2.0calc.com/questions/help-pls-fast_1

Rational numbers can also be negative. For example, -2/5 is a rational number, because it is expressed as the ratio of -2 to 5. 

A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero,

Rational numbers include 3/4, 5.6, 5/7

They don't include, 3/0, 4/0, and pi.....

Here is a link to the above function:

https://www.desmos.com/calculator/film7nngy4

Basically, a rational number is anything that can be written as a Ratio.

Numbers such as 1/2, 1/3, 0.10.

Hello, how whould I get cosb and c then?

Use the Cosine rule: \(\cos A = \frac{b^2+c^2-a^2}{2bc}\)

Here:  a = 8, b = 7, c = 5

Cycle a, b and c for the other angles.

What are the areas of triangles ABC and DEF, in that order?

Hello Guest!

\(a=30^\circ\\ h_1=5\cdot sin\ 30^\circ=2.5\\ \color{blue}A_{ABC}=\dfrac{4\cdot h_1}{2}=\dfrac{4\cdot 2.5}{2}=5\)

\(d=arccos\ 0.6\\ h_2=5\cdot sin\ arccos\ 0.6=4\\ \color{blue}A_{DEF}=\dfrac{4\cdot h_2}{2}=\dfrac{4\cdot 4}{2}=8\)

  !

Thanks Builderboi,   

Here ia another approach. 

\(\left(\frac{9}{25}\right)^{-\frac{3}{2}}-\frac{\sqrt x}{27}=\left(\frac{3^3}{11^2}\right )^{-1}\\~\\ \left(\frac{3}{5}\right)^{-3}-\frac{\sqrt x}{27}=\left(\frac{27}{121}\right )^{-1}\\~\\ \left(\frac{5}{3}\right)^{3}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ \frac{125}{27}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ 125-\sqrt x = 121\\~\\ \sqrt x = 4\\~\\ x=16 \)

The composite numbers whose prime factors fall between 30 & 140 are 961, 1147, 1271, 1333, 1369, 1457, 1517, 1591, 1643, 1681, 1739, 1763, 1829, 1849, 1891, 1927, 1961, 2021, 2077, 2173, 2183, 2201, 2209, 2257, 2263, 2279, 2419, 2449, 2479, 2491, 2501, 2537, 2573, 2623, 2627, 2701, 2747, 2759, 2773, 2809, 2867, 2881, 2911, 2923, 2993, 3007, 3053, 3071, 3127, 3131, 3139, 3149, 3193, 3233, 3239, 3293, 3317, 3337, 3379, 3397, 3403, 3431, 3481, 3503, 3551, 3569, 3589, 3599, 3649, 3713, 3721, 3737, 3763, 3811, 3827, 3869, 3901, 3953, 3977, 4087, 4141, 4171, 4181, 4183, 4187, 4189, 4307, 4331, 4343, 4399, 4453, 4489, 4559, 4661, 4717, 4747, 4757, 4819, 4891, 4897, 5041, 5063, 5141, 5143, 5183, 5251, 5293, 5329, 5353, 5429, 5459, 5461, 5561, 5609, 5671, 5723, 5767, 5893, 5917, 5959, 5963, 6059, 6161, 6241, 6319, 6497, 6499, 6557, 6767, 6887, 6889, 7031, 7171, 7081, 7387, 7373, 7663, 7921, 7979, 8051, 8249, 8383, 8633, 8927, 8989, 9379, 9409, 9563, 9797, 9991, 10001, 10201, 10403, 10609, 11021, 11227, 11449, 11881 and 19321

They ALL have prime factors factored between these prime numbers,

31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137 & 139

After you find the first root  1 + i

you can picture finding the other roots by making (around the origin)

a 90^o rotation, giving you  -1 + i

a 180^o rotation, giving you  -1 - i

a 270^o rotation, giving you  1 - i

A 360^o rotation will take you back to your original root.

There will be four fourth roots, separated by 90^o [ 360^o / 4  =  90^o ]

There will be three third roots, separated by 120^o

There will be five fifth roots, separated by 72^o

Etc.

It's up to interpretation. The problem isn't specific on what exactly it's asking for. You could interpret it as Chris did, with exactly one 1 and exactly one 2, or you could interpret it the way you did, where you need at least one 2. 

The remaining 3 are very similar to \(1 + i\). You can find them yourself by taking the inverse(s) of 1 or both terms. 

CPhill: A minor typo in this bit: 5 nCr 2*(1/6)^2*(5/6)^3 ==1250 / 6^5==625 / 3888==16.075%

There are three other cases left