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(-3-7square)to second power

(-3-7square)2nd power

Research, Knowledge and Information :


Square (algebra) - Wikipedia


Squaring is the same as raising to the power 2, and is denoted by a superscript 2; for instance, the square of 3 may be written as 3 2, which is the number 9.
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The square root and power functions - MathOnWeb


How to use the square root function in the ... The power functions with powers p and 1/p ... The problem is the second case where −3 doesn't get mapped back ...
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What is 4 times the square root of 3, to the second power ...


What is 4 times the square root of 3, to the second power? ... Second equation (4*3^½)^2. ... What is the square root of 2 to the negative 9 plus 3 power?
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What is 3 to the second power and 3 squared - Answers.com


What is 3 to the second power and 3 squared? ... 3 to the second power is 9 and 3 squared is 9. now do whatever you need to do with those nines.
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Powers on Your Computer's Calculator - Math N Stuff


Powers on Your Computer's Calculator. Your Computer's Calculator ... (-3)², -3 to the second power [=] is -27, (-3)³, -3 to the third power [=] is 81,
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What is 3 to the second power - Answers.com


The square (second power) of 6 is (6 x 6) = 36. The square of 3.1 is (3.1 x 3.1) ... 5.9x to the third power plus 3.4x to the second power - 7 parentheses ...
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Chapter 7 - EXPONENTS AND RADICALS - tpub.com


CHAPTER 7. EXPONENTS AND RADICALS. ... it is called a square root. Thus, ... 5 2 = 25 is read "5 to the second power (or 5 squared) ...
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Powers 1 - |LASP|CU-Boulder


If 3 is raised to the 2nd power, we get 9 and if 3 is raised to ... when a number to some power is raised to a power again, we multiply the two powers to get the ...
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Suggested Questions And Answer :


x to the second power minus 2 x y plus 4y to the second power

x to the second power minus 2 x y plus 4y to the second power This comes out as: x^2 - 2y + (4y)^2
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Please help solve...

Group similar powers of X together so we have -15+2+10=-3X^2. The expression now becomes 11X^3-3X^2-3X. X is a common factor: X(11X^2-3X-3). The quadratic doesn't factorise further.
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what does x second power - 4 second power= the square root of 15 to the second power equal

"square root av fifteen tu the sekond power"...15
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simplify the negative exponents 5 negative one power and 4 negative second power ?

To find the answer you just need to find the number to the power without a negative and put it under a fraction bar with 1 over it. 5 to the negative first= 1/5 and 4 to the negative second= 1/16
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How do i solve this? x^2 +5x +6 = 2

Problem: How do i solve this? x^2 +5x +6 = 2 The first number is x squared to the second power. The first number is x squared, also called x to the second power. It is NOT x squared to the second power. That would be x^4. x^2 + 5x + 6 = 2 x^2 + 5x + 6 - 2 = 2 - 2 x^2 + 5x + 4 = 0 (x + 4)(x + 1) = 0 One or both must be zero. x + 4 = 0 x = -4 x + 1 = 0 x = -1 Check. x^2 + 5x + 6 = 2 (-4)^2 + 5(-4) + 6 = 2 16 - 20 + 6 = 2 2 = 2 x^2 + 5x + 6 = 2 (-1)^2 + 5(-1) + 6 = 2 1 - 5 + 6 = 2 2 = 2 Answer: x = -4  and  x = - 1  
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1. x to the second power + 6x +9 2. 4x to the second power +20x + 25 3. 36x to the second power - 24x +16

?????????? did yu tri tu sae "x^2 +6x+9.2=0" ??????????? quadratik equashun giv roots=-3+-0.447213595i (komplex variabels) b^2-4ac=-0.8, so sqrt(-0.8) giv 0.894427191i
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what is 1 x 10 to the second power divided by 1 x 10 to the negative third power?

10^2/ 10^-3=10^2 * 10^3=10^5=1e5
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2x to the third power + 3x to the second power - 8x = 12

2x^3 + 3x^2 - 8x = 12 this is a cubic equation so set f(x) = 2x^3 + 3x^2 - 8x - 12 = 0 and search for roots. f(0) = -12,   negative f(1) = -15,   more negative f(2) = 16 + 12 - 16 - 12 = 0 -- a root has been found. (x - 2) is a factor of f(x), so divide f(x) by (x-2), giving (x - 2)(2x^2 + 7x + 6) = 0    and the quadratic here factorises to give, (x - 2)(2x + 3)(x + 2) = 0 So the solutions are: x = 2, x = -3/2, x = -2
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1.5 divided by(0-1) to the second power x 2 =

(1.5 / ((0 - 1) to the second)) x 2 = 3   Answer: 3
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EXPAND USING BINOMIAL THOREM(2X+3)^5

EXPAND USING BINOMIAL THOREM(2X+3)^5 I am not sure what you are looking for, but I believe you need to recognize that the power is 5 and the two terms needed are 2x and 3. To expand take the first term and put it to power 5 and the second term to power 1. The steps are to decrease the first termand increase the second term. (2x)^5 + 3 + (2x)^4 + 3^2 + (2x)^3 + 3^3 + (2x)^2 + 3^4 + 2x + 3^5 = 32x^5 + 3 + 16x^4 + 9 + 8x^3 + 27 + 4x^2 + 64 + 2x + 192 = 32x^5 + 16x^4 + 8x^3 + 4x^2 + 2x + (3 + 9 + 27 + 64 + 192 = 32x^5 + 16x^4 + 8x^3 + 4x^2 + 2x + 295
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