What is the 36th power of 2?

What is the 36th power of 2?

2^36
=16^9
=4096^3
=68719476736

If a + B power of x = 64, B power of x = 2, find a power of X

x^(a+b)=x^a*x^b
Because a + B power of x = 64, B power of x = 2
That is 64 = 2 * x ^ a
The solution is x ^ a = 32

Calculate the 5th power of 2 times the 5th power of 4

The fifth power of 2 times the fifth power of 4
=The 5th power of 2 multiplied by the 5th * 2nd power of 2
=5 + 5 * 2 power of 2
=The 15th power of 2

11 9 4 1 - how to share with others? 14 20 15 3

The least common multiple of these four numbers is 11 × 9 × 4 = 396
So 396 is their denominator
The least common multiple of these four numbers is: 4 × 5 × 3 × 7 = 420
So 420 is their denominator
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() / () = () / () = () / () = 3, 1-9 numbers are filled in brackets to make the formula true
It can't be repeated

Because the number is selected from 1-9, and the result is equal to 3, so the molecule naturally can not be greater than 3 (because 3 * 4 is already 12 and greater than 9), so the molecule can only be 1, 2, 3 respectively, and the denominator can only be corresponding to 3, 6, 9, but 3 will be repeated, so you can only say that you have a problem, there should be no (can not be repeated) exact answer

Given that the minimum value of quadratic function y = x & sup2; + (2k + 1) x + K & sup2; - 1 is 0, the value of K can be obtained
It takes some rough process

The minimum value of this function is 0 = > only one root = > (2k + 1) ^ 2-4 (k ^ 2-1) = 0
The solution is k = - 5 / 4

In 1 to 100 integers, find 10 numbers, is their reciprocal sum equal to 1, please

2… 6… 10… 12… 20… 30… 42… 56… 72… ninety

Calculation: 1. 4 / 5 × 3 / 4 × 2 / 9, 4 / 3 × 12 × 1 / 2, 26 / 7 × 2 / 7 × 13
Fill in the blank number with ＞, < or =
3 / 4 × 2 / 3 〔 3 / 4 × 3 / 4 × 5 / 4 〕 3 / 4

1, four fifths × three fourths × two ninths
=4/5*3/4*2/9
=2/15
2. 4 / 3 × 12 × 1 / 2
=3/4*12*1/2
=9/2
3. 7 / 26 × 2 / 7 × 13
=7/26*2/7*13
=1
Fill in the blank number with ＞, < or =
3 / 4 × 2 / 3 [<] 3 / 4 × 5 / 4 [<] 3 / 4

How to calculate simply 2001 times [1 / 8 minus 1 / 2009] plus 8 times [1 / 2001 minus 1 / 2009] minus 2009 times [1 / 8 plus 1 / 2001] plus 8
Experts help, the answer is good reward

2001*(1/8-1/2009)+8*(1/2001-1/2009)-2009*(1/8+1/2001)+8=2001/8-2001/2009+8/2001-8/2009-2009/8-2009/2001+8=2001/8-2009/8-2001/2009-8/2009-2009/2001+8/2001+8=(2001-2009)/8-(2001+8)/2009-(2009-8)/2001+8=...

It is known that the root of the quadratic equation AX & # 178; + BX + C = 0 (a is not equal to 0) is - 1 and a = √ 1 / 2 (C-2) + √ 1 / 2 (2-C) - 3,
Finding the power of cb2007 + (C-2) 2008

If the root sign is significant, then 1 / 2 (X-2) > = 0, C > = 2
1/2(2-c)?=0,c