Guide :

# 0=[22.12848]2 to the power times a times 81.4m

gswqydt2qfdtyqgdty3wgf

## Research, Knowledge and Information :

### Math Forum: Ask Dr. Math FAQ: N to Zero Power

What is n^0 (any number to zero power ... 3^3, 3^4, .... Finding the actual values, we get 3, 9, 27, 81 ... every time you move to the right ...

### Adjustable Frequency AC Drive - Rockwell Automation

... (2.0 HP) ratings for the 22F ... • In this manual we refer to the PowerFlex 4M Adjustable Frequency AC Drive as: ... 2. Actual deceleration times can be longer ...

### Toyota M engine - Wikipedia

... (at the same time as the 4-cylinder 18R-E). The 4M-E was the first Toyota engine to be equipped ... The first M was a 2.0 L ... 8.1:1 to 9.2:1; Power: 145–175 ...

### What is 6.02 times 10 to the 23rd power? | Reference.com

The equation of 6.02 times 10 to the 23rd power is equal to ... Exponents are numbers that indicate the number of times a function is multiplied ...

### BMD-4 - Wikipedia

BMD-4: 2V-06-2 water-cooled diesel engine ... BMD-4M: 500 hp (368 kW) Power/weight: BMD-4: 33.1 hp/tonne (24.3 kW/tonne) BMD-4M: 37.0 hp/tonne (27.2 kW/tonne) Suspension:

### What Is 18 raised to the 1/4 power times 12 raised to the 1/2 ...

0 0 What Is 18 raised to the 1/4 power times 12 raised to the 1/2 power. I need to be shown how to do this please please please!! 2/5/2013 ...

### 5011656403-4M03 クイックスタート Quick Start

5.5 (0.22) g h i j k ... (Blinks 3 times) Apply Power ... 0 PowerFlex 4M ' 120 mm (4.7 in.) 120 mm (4.7 in.) Closest object that may obstruct air flow

### The Binomial Theorem: Examples - Purplemath | Home

Demonstrates how to answer typical problems involving the Binomial Theorem. ... (x 4)(81) + (6)(x 2 ... (which always equals 1) times a n times b 0 (which ...

### Powers on Your Computer's Calculator - Math N Stuff

Powers on Your Computer's Calculator. ... (-3)³, -3 to the third power [=] is 81,-3 to the fourth power [=] ... A Number Raised to 0 Power is One, ...

## Suggested Questions And Answer :

### x to the 3rd power y to the 4th power times xy to the 2nd power

x^3y^4 times (xy)^2=x^5y^6

### 20 times x to the negative 4th power divided by 27 times y to the 2nd power

(20x^-4) / (27y^2) =(20/27) / x^4 y^2 or 0.740740...*x^-4 y^-2

### formula used to solve this equation: 5power of x+1 times 25power of x equals 5 power of 2x-3

(5^(x+1)(25^x)=5^(2x-3). 25=5^2, so 25^x=5^(2x). Now everything is in terms of powers of 5. We can combine the multiplication on the left: 5^(x+1+2x)=5^(2x-3). The power of 5 on the left matches the power of 5 on the right so 3x+1=2x-3 from which x=-4. Check by substituting for x in the original equation: 5^-3*25^-4=5^-3*(5*5)^-4=5^-3*5^-4*5^-4=5^(-3-4-4)=5^-11. On the right: 5^(-8-3)=5^-11. Checks out OK!

### Find the Speed of each boat

Question: It takes a Cabin Cruiser 5 hours longer to travel 68 Kms then it takes a power boat to travel 102 kms. The Cuiser travels one Quarter the speed of the power boat. Find the Speed of each boat. Let Vcc be the speed of the Cabin Cruiser. Let Vpb be the speed of the power boat. Then Vpb = 4*Vcc. Time for cabin cruiser top travel 68 km is Tcc = 68/Vcc hrs Time for power boat to travel 102 km is Tpb = 102/Vpb hrs. And Tcc = Tpb + 5. 68/Vcc = 102/Vpb + 5 68/Vcc = 102/(4Vcc) + 5 68/Vcc = 25.5/Vcc + 5 (68 - 25.5)/Vcc = 5 42.5/Vcc = 5 42.5/5 = Vcc Vcc = 8.5 km/hr Vpb = 34 km/hr

### simplify 530712 as powers of 2 and, or 3.

If you factorize 530712: 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 x 7 x 13 Which you can then rewrite as:     3        6 2 x 3 x 7 x 13 (2 to the power of 3 times 3 to the power of 6 times 7 times 13). Check your factorization here http://statsfiddle.info/Primes/PrimeFactors/530712)

### why 64 has an odd number of factors

Actually, it doesn't have odd number factors because it's base power is only 2 this is 2^6 (2 to the power of 6)= 64 which means that it is 2*2*2*2*2*2 ( 2 times 2 times 2 times 2 times 2 times 2) = 64 SO , it doesn't have an odd number factor at all

### 2/3 times 2/3 times 2/3 times 2/3 times 2/3

= (2/3)^5 {to the fifth power means multiplied by itself five times} 2^5 ----- {took numerator and denominator, both to the fifth power} 3^5 32 ----- {evaluated exponents} 243 www.algebrahouse.com

### 64239 is divided by a certain number.while dividing the number 175,114 and,213 appears as three successive remainders.find the divisor ?

It's not clear what "successive remainders" means. There is only one remainder after a division: (dividend)=(quotient)*(divisor)+(remainder). For example, if 175 is the remainder after division then 286 divides into 64239 224 times, remainder 175. The largest remainder is 213, implying that the divisor is bigger than 213. It's clear that the divisor cannot be divided again by the same divisor to yield a remainder in excess of the divisor. In the example 286 will not divide into 224, so 224 would be a second remainder. If the divisor has to be used three times to divide successively into each quotient to obtain each remainder then the divisor cannot exceed 40 (the approximate cube root of 64239) and therefore the remainders cannot exceed 40. See explanations in square brackets below. The only other interpretation I can find to the question is that in the process of long division, we have the successive remainders as we proceed with the division. But that doesn't seem to work either. [The numbers a, b and c are integers and b=cx+213; a=bx+114; 64239=ax+175, making ax=64064. Combining these equations we get: a=64064/x=bx+114, so b=64064/x^2-114/x=cx+213, and c=64064/x^3-114/x^2-213/x. Because 64064=ax, we see that 64064 must be divisible by x. 64064 factorises: 2^6*7*11*13. Because the largest remainder is 213, x>213, because x must be bigger than any remainder it creates. But the first term 64064/x^3 can only yield an integer value if x^3<64064, that is, x<40, which contradicts the requirement x>213.] [If the successive remainders appear in the long division, and the divisor has 3 digits as expected, and three remainders suggest that the other factor also has 3 digits, then if 175 is the first remainder it must be the result of dividing the divisor into 642. Therefore the multiple of the divisor =642-175=467. Since 467 is prime then the first digit of the other factor must be 1. 64239 divides by 467 137 times with 260 as the remainder. This is not one of the three remainders, so 175 cannot be the first remainder. Try 114: 642-114=528=2^4*3*11. We need a factor >213. 264, 352. This time we appear to have partial success, because 176 (less than 213, but half of 352) goes into 64239 364 times with a remainder of 175! So we have two of the three required remainders. Unfortunately the third remainder is 87, so 114 cannot be the first remainder. Finally, try 213: 642-213=429=3*11*13. The only factor bigger than 213 is 429 itself, and the final remainder after dividing 64239 by 429 is 318, not included in the three remainders given. So the assumption that the remainders arise during long division is also false.] 64064=2^6*7*11*13. Leaving aside the powers of 2, and just looking at combinations of 7, 11 and 13, we get 7*11=77; 7*13=91; 11*13=143, and 7*11*13=1001. By progressively doubling these products we can discover factors bigger than 213: 308=4*77; 364=4*91; 286=2*143, etc. We can also take 7, 11 and 13 and multiply them by powers of 2: 224=32*7; 352=32*11; 416=32*13. We can write the factors of 64064 in pairs, starting with a factor a little bigger than 213: (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104), (704,91), (1001,64), (1232,52), (2002,32), (2464,26), (4004,16), (4928,13), (8008,8), (16016,4), (32032,2). These are (a,x) or (x,a) pairs. If we take 175 as the remainder, and ignore the other two, then we have a choice of factors. The 3-digit factors are (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104). The only pair in which each factor is greater than 213 is (224,286).

### what is the simplified version of 2m to the second power times 2m to the power of three?

2m^2 * 2m^3=4m^5 OR (2m)^2 * (2m)^3=4m^2 * 8m^3=32m^5 Depends on interpretation