Evaluate 27^2\27 ^4/3

2 answers:

7 0

Following the rules of PEMDAS:

We would do exponents before the division

27 ^ 2 = 729

27 ^ 4 = 531,441

729/531441/3 = 0.000456247 approximately

NISA [10]3 years ago

7 0

Answer: Its 9.

Step-by-step explanation:

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the class marks of a distribution are 37 42 and 47 the class limits corresponding to class mark 42 are

I am Lyosha [343]

Answer:

$39.5,44.5$

Step-by-step explanation:

$Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ first\ interval\ be\ u_1,v_1.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ second\ interval\ be\ u_2,v_2.\\Let\ the\ lower\ limit\ and\ the\ upper\ limit\ of\ the\ third\ interval\ be\ u_3,v_3.\\Hence,\\As\ the\ class\ marks\ are\ uniform\ with\ an\ a\ difference\ of\ 5, the\\ observation\ is\ continuous\ and\ the\ class\ size\ is\ 5\ too.\\Hence,\\v_1=u_2, v_2=u_3\\$ $Now,\\Lets\ consider\ the\ first\ two\ classes.\\37=\frac{u_1+v_1}{2} \\42=\frac{u_2+v_2}{2}\\By\ adding\ the\ equations:\\79= \frac{u_1+v_1}{2}+\frac{u_2+v_2}{2}\\79=\frac{u_1+v_1+u_2+v_2}{2}\\79=\frac{(u_1+v_2)+(v_1+u_2)}{2}\\79=\frac{(u_1+v_2)}{2}+\frac{(v_1+u_2)}{2}\\Now,\\As\ the\ mid-point\ between\ two\ points\ can\ be\ calculated\ by\\ its\ average:\frac{u+v}{2} \\Hence,\\u_2\ lies\ in\ the\ mid-point\ of\ u_1\ and\ v_2.\\u_2=\frac{(u_1+v_2)}{2}\\$

$Hence,\\By\ substituting\ u_2=\frac{(u_1+v_2)}{2},\\79=u_2+\frac{v_1+u_2}{2}\\As\ v_1=u_2[Proven],\\79=u_2+ \frac{u_2+u_2}{2}\\79=u_2+\frac{2u_2}{2}\\79=2u_2\\Hence,\\u_2=\frac{79}{2}\\u_2=39.5\\Now,\ as\ we\ already\ know\ that\ the\ class\ size=5,\\v_2-u_2=5\\Hence,\\v_2=5+u_2\\Here,\\v_2=5+39.5\\v_2=44.5\\Hence,\\The\ upper\ limit\ of\ the\ second\ interval=44.5\\The\ lower\ limit\ of\ the\ second\ interval=39.5\\$

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3 years ago

Put the following equation of a line into slope-intercept form, simplifying all

Charra [1.4K]

4x + 2y = -18

2x + y = -9

y = -2x - 9

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3 years ago

3. WEATHER On a cold January day,

erma4kov [3.2K]

Answer: -9 + 21 = t t = 12°C.

Step-by-step explanation: For this problem lets have t be equal to the original temperature in the beginning of the day. The equation -9 + 21 = t represents this situation because it states that if we add the degrees that dropped to the current degrees it will give us our original degrees.

Solve.

-9 + 21 = t (Simply add -9 + 21) -9 + 21 = (+12)

12 = t

So, the original degrees in the beginning of the day was 12°C.

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3 years ago

(a + b)(a 2 - ab + b 2)

JulsSmile [24]

Answer:

a³ + b³

Step-by-step explanation:

Given