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# Can someone help me with sqaure numbers? Topic: Can someone help me with sqaure numbers?
December 13, 2019 / By Bashemath
Question: I don't fully understand what i square number is, ive tried google first but i am one of those people that need it explaining quite a bit to understand, say for example if i had a question on a maths paper saying what is the square number of 9, would the answer be 3 as 3x3 is 9? please help!! Thanks ## Best Answers: Can someone help me with sqaure numbers? Adelice | 10 days ago
Hi square numbers are which make the number such as 1, 4, 9, 16 etc. these would be gained by 1*1=1 2*2=4 etc the square of 9 = 9*9= 81 so the answer to that is 81 however you seem to know the square root- which in simplest terms means that a number lets say 16 which needs a square number required to make that which is 4 it would make it easier by wrting all the square numbers out to make it easier, this applies for cube numbers and cube roots hope i helped :D
👍 208 | 👎 10
Did you like the answer? Can someone help me with sqaure numbers? Share with your friends Originally Answered: Complex Numbers, Real Numbers, & Imaginarty Numbers?
yahoo seems to be filtering the math. complex numbers follow the rules of normal mathematics, except i*i=-1 which may seem a bit odd. If you want to visually repressent them, you could start out with a normal x-y graph, and replace the y axis with the i. and then plot your points. you can express complex numbers in two formats. one is in the rectangular form a+b*i, the latter is the r*e^(i*theta). r*e^(theta*i) can also be expressed as r*(cos(theta*i)+i*sin(theta*i)) or r*cis(theta*i) for short. the rectangular form may be obvious. the polar form is you want to find the length of a line segment from the origin to the point called the modulus which basically is the pythagorian theorem, and also the angle it makes bewteen the range of 0 and 2pi radians. just make sure your angle is the real angel mapped to the quadrant it is in, and not just what your calculator returns. personally I would use the rectangular form if I am adding, subtracting or multiplying complex numbers. however, i would use the polar form if i were to divide, or raise a complex number to a power-also it's pretty fast in multiplication too. if you have two numbers a+bi and c+di their sum is: (a+c)+(b+d)*i their difference is (a-c)+(b-d)*i. their product is: (a*c-b*d)+(ad+b*c)*i their division is not something easy to remember: (a*c+b*d)/(c*c+d*d)+i*(b*c-a*d)/(c*c+d... also if you have a*cis(c) and b*cis(d) then their sum is a*cos(c)+b*cos(d)+i*(a*sin(c)+b*sin(d)... difference is a*cos(c)-b*cos(d)+i*(a*sin(c)-b*sin(d)... product: remember cis(theta)=e^(theta*i) so using e, we can add the exponents: a*b*e^(i*(c+d)) a*b*cis(i*(c+d)) and division: (a/b)*e^(i*(c-d)) (a/b)*cis(i*(c-d)) there are proofs using trig, but using the exponential is must easier. to raise powers: a*cis(theta)^n a^n*cis(n*theta) however if n is expressed as 1/k where k is a positive integer and not zero it has k possible n roots. all these roots are points along a circle with a radius of r^(1/k), connecting the points will form a normal n-gon or k-gon in this case. a^k*cis((2*pi/k)*j) where j is an integer. you may recall: k^m=n and ln(n)/ln(k)=m then ln(n)=m*ln(k) => n=e^(m*ln(k)) so it follows that if you have (a+b*i)^(c+d*I)=some number call x then x=e^((c+d*i)*ln(a+b*i)) with ln(a+b*i)=ln(a^2+b^2)/2+(pi/2*sign(b)-at... so x=e((c+d*i)*(ln(a^2+b^2)/2+(pi/2*sign(... foil and you get some e^(real+imaginary)=e^real*e^(imaginary [i implied]) with real=c*ln(r)+d*(atan(a/b)-sign(b)*pi/2) and imaginary=d*ln(r)-c*(atan(a/b)-sign(b)*p... another method, my parentesee may be messed up: (a+b*i)^(c+d*i) could be expressed as r*cis(i*theta)^c*(r*cis(i*theta))^(d*i... r*e^(i*theta)^c*(r*e^(i*theta))^(d*i) r*e^(i*theta)^c*e^ln((r*e^(i*theta))^(... r*e^(i*theta)^c*e^(d*i*ln(r*e^(i*theta... r*e^(i*theta)^c*e^(d*i*(ln(r)+ln(e^(i*... r*e^(i*theta)^c*e^(d*i*(ln(r)+i*theta(... r*e^(i*theta)^c*e^(d*i*ln(r)+d*i*i*the... r^c*e^(i*c*theta)*e^(d*i*ln(r)-d*theta... r^c*e^(i*c*theta)*e^(d*i*ln(r)) /(e^(d*theta)) Teagan
First off, a square number is a square root. So the square root of 9 is 3, because 3 times it's self is 9. 3 squared, means 3x3. 3 cubed means 3x3x3. And so on, so 7 to the power of 6, is 7x7x7x7x7x7. What ever the small number in the right hand corner of a number, is called "to the power of" or an "exponent". What ever the exponent is, is how many times you multiply the number, that is the large one. So let's say the number in <5> is a small number, or an exponent. 3<5> = 3x3x3x3x3 (5 threes)
👍 80 | 👎 9 Peyton
The square root of a number means the GCF of a number, written as a power (or an exponent). For instance, the square root of 9= 3² or 81=9².
👍 71 | 👎 8 Lyle
When you multiply a number by itself, that number is squared. So 3×3=9 means that 9 is 3 squared. Also, 3 is the square root of 9.
👍 62 | 👎 7 Jefferson
The square of a number is it it times its self a squared would be a*a 9 squared would be 9*9 so 81 but you were talking about a square root it is the opposite of square square root of 81 would be 9*9=81 so 9 and the square root of 9 would be 3 hope this helps
👍 53 | 👎 6 Garnet
A square number is any number (whole, fraction, decimal) multiplied by itself. ex. 2x2=4 8x8=64 12.3x12.3=159.29
👍 44 | 👎 5 Originally Answered: The Difference between two positive numbers is 5, and their product is 84. Find the numbers?
Here, we have... x - y = 5 xy = 84 For this problem, it's easier to use substitution method... Here is how I use it... x = y + 5 [Solve for x for the first expression] y(y + 5) = 84 [Substitute the first expression for the second and solve for y] y² + 5y = 84 y² + 5y - 84 = 0 (y - 7)(y + 12) = 0 y = 7 [y = -12 is rejected since you are determining positive value.] Hence... 7 * x = 84 x = 12 The values are 12 and 7. I hope this helps!

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