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3 ft 10 in = 2 ft 22 inches

2 ft 22 inches - 11 inches = 2ft 11 inches

(400)/ [(400^2 - (400-20)(400+20) ] =

400(400^2 -400^2 +20^2) =

400/(20^2)

400/400 = 1     No calculator required !

Two thousand three hundred and thirty three.

In mixed number form it is 1 1/5. If you need it simply in a fraction you can also write it as 6/5.

1 2/10 

simplified is: 1 1/5

Nvm I know how to do liao

So, what is the question?

thats awful

These are exact values!  Presumably, you mean that, given they are angles, what are their values in degrees?

¶ radians = 180° so

1. 5¶/3 radians = 5*180/3 = 300°

etc.

1.2 = 1 1/5

1 minute = 60 seconds,
60*60 = 3600. //That's the number of seconds in an hour.
3600* 24 = 86400 // number of seconds in a day.
86400*365 = 31 536 000 //Number of seconds in a year.
 

look it up

google is your best friend

The equation noted by Gauss should be of course:

\(\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\cdot \left[~ \sqrt x ~\right] + 4 \cdot \left[~ \sqrt{\frac{x}{2}} ~\right]^2 + \ 8 \sum \limits_{y_1=\left[~ \sqrt{\frac{x}{2}} ~\right]+1}^{\left[~ \sqrt x ~\right]} \left[~ \sqrt{x-y_1^2} ~\right]\\ \hline \end{array}\)

Calculation of the Number of Lattice Points in the Circle

Let \(r\) be the Radius of the Circle

Let \(x = r^2\)

Let \( A_2(x)\) be the number of lattice points in circle

Let \([\ldots] \) be the largest integer

Noted by Gauss:

\(\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\cdot \left[~ \sqrt x ~\right] + 4 \cdot \left[~ \sqrt{\frac{x}{2}} ~\right]^2 + \sum \limits_{y_1=\left[~ \sqrt{\frac{x}{2}} ~\right]+1}^{\left[~ \sqrt x ~\right]} \left[~ \sqrt{x-y_1^2} ~\right]\\ \hline \end{array}\)

Computed Results

\(\begin{array}{r|r} \hline r = \sqrt{x} & A_2(x) \\ \hline 1&5\\ 2&13\\ 3&29\\ 4&49\\ 5&81\\ 6&113\\ 7&149\\ 8&197\\ 9&253\\ 10&317\\ 11&377\\ 12&441\\ 13&529\\ 14&613\\ 15&709\\ 16&797\\ 17&901\\ 18&1009\\ 19&1129\\ 20&1257\\ 21&1373\\ 22&1517\\ 23&1653\\ 24&1793\\ 25&1961\\ 26&2121\\ 27&2289\\ 28&2453\\ 29&2629\\ 30&2821\\ 31&3001\\ 32&3209\\ 33&3409\\ 34&3625\\ 35&3853\\ 36&4053\\ 37&4293\\ 38&4513\\ 39&4777\\ 40&5025\\ 41&5261\\ 42&5525\\ 43&5789\\ 44&6077\\ 45&6361\\ 46&6625\\ 47&6921\\ 48&7213\\ 49&7525\\ \hline \end{array}\)

Continued

\(\begin{array}{r|r} \hline r = \sqrt{x} & A_2(x) \\ \hline 50&7845\\ 51&8173\\ 52&8497\\ 53&8809\\ 54&9145\\ 55&9477\\ 56&9845\\ 57&10189\\ 58&10557\\ 59&10913\\ 60&11289\\ 61&11681\\ 62&12061\\ 63&12453\\ 64&12853\\ 65&13273\\ 66&13673\\ 67&14073\\ 68&14505\\ 69&14949\\ 70&15373\\ 71&15813\\ 72&16241\\ 73&16729\\ 74&17193\\ 75&17665\\ 76&18125\\ 77&18605\\ 78&19109\\ 79&19577\\ 80&20081\\ 81&20593\\ 82&21101\\ 83&21629\\ 84&22133\\ 85&22701\\ 86&23217\\ 87&23769\\ 88&24313\\ 89&24845\\ 90&25445\\ 91&25997\\ 92&26565\\ 93&27145\\ 94&27729\\ 95&28345\\ 96&28917\\ 97&29525\\ 98&30149\\ 99&30757\\ 100&31417\\ \hline \end{array}\)

6/5x1/2= 3/5 (0.6)

ʃ (2x + 6)(x2 + 6x + 3)7 dx

Sorry, without mistakes:

\(\begin{array}{rcll} \hline && \int (2x + 6)(x^2 + 6x + 3)\cdot 7\ dx \\ \hline \end{array}\)

\(\begin{array}{rcll} &=& 7\cdot \int (2x + 6)(x^2 + 6x + 3)\ dx \\ &=& 7\cdot \int (2x^3+12x^2+6x+6x^2+36x+18)\ dx \\ &=& 7\cdot \int (2x^3+18x^2+42x+18)\ dx \\ &=& 7\cdot \left( \frac24x^4+\frac{18}{3}x^3+\frac{42}{2}x^2 + 18x \right) \\ &=& 3.5x^4+42x^3+147x^2+126x + c \\ \end{array}\)

g=6-7n/3     Subtract 6 from both sides of the equation

g-6 = -7n/3   Multiply both sides by  - 3/7

-3/7 (g-6) = n

-3/7 g + 18/7 = n

Using Web2.0 calculator:

21*11-33*2 = 165

f and g differ between -1<x<1:

.

Do you have more information?   asin(.5) = 30   or 150 degrees..... are you sure you calculated  0.5  as the value for which you need the arcsin?