Wednesday

April 16, 2014

April 16, 2014

Number of results: 19,127

**Math**

Real numbers include rational, irrational, integers, and natural numbers. Real numbers form a subset of complex numbers. So when a number is not real, it is usually complex. Can you spot a complex number from the list?
*Thursday, October 28, 2010 at 7:33pm by MathMate*

**math-complex numbers**

Sketch the sets of complex numbers in the complex plane that satisfy the following: a) |z-1|=2 b) |z-3i|=2 c) |z-5+2i|=2 I'm not sure, but are you supposed to rearrange the whole thing to |z|=a+bi? So, part b would be |z|=2+3i.
*Wednesday, October 9, 2013 at 9:34pm by Rin-chan-san*

**Math**

I am in the 5th grade and my homework tonight is about complex numbers. I have no idea how to do complex numbers, not to mention, I dont know what they are. Can you help me?
*Wednesday, October 27, 2010 at 3:42pm by Jeida *

**Complex Numbers**

Oh thank you so much i forgot that when you that the square root of a negative you get the complex #
*Monday, January 31, 2011 at 9:03pm by Allie*

**Complex numbers - quadratic formula**

Let a and b be real numbers. The complex number 4 - 5i is a root of the quadratic z^2 + (a + 8i) z + (-39 + bi) = 0. What is the other root? I did a lot of work on hand and with wolfram alpha but it doesn't simplify itself easily at all....anyone have any help to offer? Thanks
*Thursday, January 9, 2014 at 11:58am by Cambridge*

**Calculus--complex numbers**

the complex number Z satisfies Z/(Z+4)= 3+2i Find Z. please help thx!
*Thursday, March 13, 2008 at 7:11am by Anne*

**complex numbers**

the complex number Z satisfies Z+i/(1+2i)= iz Find Z. please help thx!
*Tuesday, November 4, 2008 at 7:22am by Mary*

**complex numbers**

the complex number Z satisfies (Z+i)/(1+2i)= iz Find Z. please help thx!
*Tuesday, November 4, 2008 at 1:43pm by Mary*

**Math (Complex Numbers)**

Let a,b,c be complex numbers satisfying a+b+c=abc=1 and (ab+bc+ac)/3=(1/a^2)+(1/b^2)+(1c^2) The sum of absolute values of all possible ab+bc+ac can be written as (√n)/m, where n and m are positive coprime integers. What is n+m?
*Thursday, May 9, 2013 at 9:55am by Mahler*

**Math Help**

Is 3 sqrt -16 supposed to mean (sqrt 3) -16 or 3*(sqrt -16) ?? If you are talking about the square root of negative numbers, the answer will be a complex number. Are you studying complex numbers?
*Friday, October 19, 2007 at 7:54pm by drwls*

**algebra**

a.natural numbers Numbers that are non-negative integers, such as 1,2,12,201,... There are two definitions of natural numbers, one that includes 0, and the other one does not. Most high-school definitions include 0. b.whole numbers Whole numbers do not have decimals nor ...
*Tuesday, September 1, 2009 at 10:20pm by MathMate*

**English**

1) complex 2) complex 3) complex 4)simple 5)complex 6)complex 7)compound 8) Simple 9)simple 10)simple 11)Complex 12)complex 13)Simple
*Tuesday, November 9, 2010 at 2:10pm by Tom Jameson*

**math-complex numbers**

If z = -3+4i, determine the following related complex numbers. a)vector z b)3(vector z) c) 1/(vector z) d) |z| e) |vector z| f) (vector z)/(|z|^2)
*Friday, October 11, 2013 at 10:45am by Rinchan*

**Algebra**

2k^2 = -144 k^2 = -72 Unless you have studied complex numbers you would stop here and say "there is no real solution" since you cannot take the square root of a negative. If you studied complex numbers you should know that i = √-1 so x = √-72 x = ±6√2 i
*Monday, March 17, 2008 at 10:21pm by Reiny*

**trig**

Are you familar with the method of changing a complex number of the a + bi into polar form. If not, here is a good page http://www.intmath.com/Complex-numbers/4_Polar-form.php notice that a + bi would be a unique position on the grid, so ....
*Monday, March 1, 2010 at 4:47pm by Reiny*

**calculus**

can you please explain tep by step how to solve dividing complex numbers strictly using a graph, the complex plane. I know how to do it algebraically, but not graphically. thankyou
*Saturday, June 9, 2012 at 10:34am by wendell*

**Complex numbers**

Prove algebraically that |w+z| is less than or equal to |w|+|z| for any complex numbers w and z, where || is the magnitude. After letting w = a+bi and z = c + di and doing some plugging in etc. I got that ac-bd <= b^2 + sqrt (a^2+b^2)(c^2+d^2) + d^2 But is my proof complete...
*Thursday, January 2, 2014 at 11:24am by Cambridge*

**Algebra II**

Yes, the numbers in a matrix can have any value, positive or negative, fractions, decimals and irrational numbers, not just integers from 0 to 9. They can also be complex numbers like 4 + 3.1 i [3 4] +[2 7] = [5 11] , for example
*Saturday, October 27, 2007 at 7:25pm by drwls*

**math **

I am not sure what you mean by the "trig form" of a complex number. The magnitude is sqrt(7^2 + 2^2) = sqrt53 The complex unit vector is e^(i*theta) where theta = arcsin(-2/7) - -16.6 degrees or -0.29 radians Thus 7 -2i = sqrt53*e^-0.29i In the second questiuon, you seem to be...
*Thursday, December 30, 2010 at 11:00am by drwls*

**12th grade - Math**

The magnitude of a complex number is defined in the complex space where the real numbers are on the horizontal axis, and the complex (imaginary) part is on the vertical axis. The magnitude is defined to be the distance of the given point to the origin, namely for a complex ...
*Wednesday, September 1, 2010 at 8:22am by MathMate*

**Maths- complex numbers**

Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1 - tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1 - tan theta)^2] Then write down the equation for tan (2 theta + ...
*Tuesday, February 13, 2007 at 2:45am by Jake*

**Abstract Algebra**

Let H={a+bi a,b is a element R, a^2+b^2=1} be a subset of the non zero complex numbers C*. Prove that H<C* under complex multiplication using the one-step subgroup test. Also describe the elements of H geometrically.
*Friday, December 7, 2007 at 9:22am by Devin*

**ALGEBRA**

How many ordered triples of complex numbers(a,b,c) are there such that a^3-b,b^3-c,c^3-a are rational numbers, and a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) b}+{(b^3)c}+{(c^3)a}]
*Friday, October 11, 2013 at 8:08am by boss*

**math**

How many ordered triples of complex numbers(a,b,c) are there such that a^3- b,b^3-c,c^3-a are rational numbers, and a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) b}+{(b^3)c}+{(c^3)a}]
*Friday, October 11, 2013 at 2:43pm by frm*

**algebra ll-imaginary numbers**

do a google search for "complex numbers"
*Wednesday, March 18, 2009 at 7:50am by Reiny*

**division of complex numbers**

no, use 2-3i, because (2+3i)(2-3i) = 4 - 9i^2 = 4+9 = 13 and you get rid of the complex number in the bottom
*Monday, January 16, 2012 at 4:13pm by Steve*

**Algebra**

3 i or 3 times the square root of -1 If you have not had imaginary numbers in your course, the answer is no real number solution no, square root of 2 or [ sqrt 2 } Yes there is an answer to the square root of a negative number but not in real numbers. If you have not covered ...
*Saturday, September 8, 2012 at 8:38pm by Damon*

**algebra, math**

How many ordered triples of complex numbers(a,b,c) are there such that a^3- b,b^3-c,c^3-a are rational numbers, and a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) b}+{(b^3)c}+{(c^3)a}]
*Friday, October 11, 2013 at 1:17pm by pls help quickly *

**math**

How many ordered triples of complex numbers(a,b,c) are there such that a^3- b,b^3-c,c^3-a are rational numbers, and a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) b}+{(b^3)c}+{(c^3)a}]
*Friday, October 11, 2013 at 8:19pm by plss answer fast*

**ALGEBRA!!**

calculate P(i) for the following polynomial p(x)=x3+2x2-5x+8 Find | z | for each of the following complex numbers and represent them on a complex plane. a. z=-3-4i b. v=5-12i c.u=7+24i d.w=-9+40i e.p=1+3i f.t=-2i g.n=4 h.m=0 i.s=(√3-i)2
*Thursday, October 7, 2010 at 2:09pm by natali*

**math**

use quadratic equation x = [ -4 +/- sqrt(16 - 24) ]/6 the solution is a complex number because b^2-4ac is negative. Given your last question, I do not think you have covered complex numbers. Perhaps you have a sign error.
*Tuesday, March 4, 2014 at 3:12pm by Damon*

**math**

solve and simplify the answer completely by using pure imaginary numbers/complex numbers. 2x^2+3x+2=0
*Thursday, December 4, 2008 at 9:20pm by anonymous*

**math**

You asked this question before. I see no reason why any numbers, even complex numbers, cannot be arranged in a two-column array.
*Monday, September 10, 2007 at 4:15pm by drwls*

**oops - math **

silly me, forgot to include the link http://www.intmath.com/complex-numbers/4-polar-form.php
*Thursday, December 30, 2010 at 11:00am by Reiny*

**Pre-Calculus**

You're right, it is a W-shape, and in this case, it is more like a flattened parabola. The curve never touched the x-axis, so there are no real roots. In other words, all four roots are complex. Are you familiar with solving a quartic? and working with complex numbers?
*Tuesday, September 8, 2009 at 10:22pm by MathMate*

**English**

Choose the term that describes the type of sentence shown. If you are in line early, you'll be sure to get the best tickets. A.) compound-complex B.) complex C.) compound D.) simple I think it is C? John was late to dinner and I was late to the movies. A.) simple B.) complex C...
*Monday, December 16, 2013 at 5:07pm by Cassie*

**English**

Identify the structure used in each of the sentences below. 1. The hamburger came from Hamburg, Germany, and the hot dog came from Frankfurt. simple compound complex compound-complex 2. The idea of placing meat on a bun, however, came from the United States. simple compound ...
*Friday, December 28, 2012 at 9:04am by Fate*

**pre-calculus**

You are given 2 real roots and one complex. As I noted before, complex numbers come in pairs, so there has to be another one, 1+2i so we have f(x) = (x+1)(x-2)(x-1-2i)(x-1+2i) multiply the last two brackets, the i's will drop out
*Saturday, March 13, 2010 at 4:18pm by Reiny*

**Algebra 2**

If we restrict ourselves to positive integers, then subtraction does not have a closure property, for example: 5-7=-2 ∉ N. If we are dealing with real numbers, division does not have closure property, because we cannot divide by zero. On the other hand, non-zero real ...
*Saturday, September 3, 2011 at 10:27am by MathMate*

**algebra II**

Odd question. a quadratic always has two solutions, real, or complex. The vertex can tell you if the solutions are real (sometimes a double solution, if the vertex is on the coordinate), or complex numbers.
*Sunday, July 3, 2011 at 2:49pm by bobpursley*

**Calculus**

First, to simplify things, observe that arctan can be expressed as a logarithm in terms of complex numbers. Suppose we have two different roots (complex or real), y1 and y2. Then: ax^2 + bx + c = A(x-y1)(x-y2) for some contant A We have: 1/[(x-y1)(x-y2)] = p[1/(x-y1) - 1/(x-y2...
*Thursday, January 10, 2008 at 2:41pm by Count Iblis*

**algebra2**

an equation to degree n, has n roots. This is always true, never violated. Never. Now, if the range of the variables are over the Real Number set only, then some of the roots cannot be written because they exclude the actual roots, complex numbers. Example(X^2+1)=0 x^2=-1 x...
*Wednesday, February 23, 2011 at 2:33pm by bobpursley*

**Math**

It looks like you have not learned about imaginary numbers or complex numbers by definition √-1 = i so -√-76 + √-125 = -√36√2√-1 + √25√5√-1 = -6√2 i + 5√5 i yes, the answer does not exist in the set of real ...
*Wednesday, September 8, 2010 at 2:56pm by Reiny*

**ALGEBRA**

How many ordered triples of complex numbers(a,b,c) are there such that a^3-b,b^3-c,c^3-a are rational numbers, and a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) b}+{(b^3)c}+{(c^3)a}]
*Friday, October 11, 2013 at 10:54am by algebra,math*

**Math Help Please**

There is a little rule of thumb that you can count on for finding zeroes of polynomials with complex roots. Complex roots always come in pairs. Each root of the pair is the complex conjugate of the other. For example, if you have a root as 4+3i, the other root must be 4-3i. If...
*Wednesday, July 22, 2009 at 11:20pm by MathMate*

**algebra2-imaginary numbers**

You can add or subtract complex numbers by treating the i as a variable and combining like terms. I am having a lot of trouble figuring out these equations with imaginary numbers. 1. (3+2i)+(7-i)= 2.(1-6i)+(2-i)= 3. (2+i)-(3+i)= 4.(4+i)-(2-i)= Can someone please walk me ...
*Tuesday, January 16, 2007 at 3:52pm by chrissy*

**Abstract Algebra**

The elements of H are the complex numbers on the unit circle and if you multiply two of those you get another complex number on the unit circle, so the product of two elements of H is another element of H.
*Friday, December 7, 2007 at 9:22am by Count Iblis*

**Math**

Factor the polynomial as the product of factors that are irreducible over the real numbers. Then write the polynomial in completely factored form involving complex nonreal or imaginary numbers. x^4 + 20x^2 -44=0
*Tuesday, January 31, 2012 at 7:38pm by Chelsea*

**maths**

There are 3 solutions with real numbers and 4 solutions with complex numbers
*Monday, April 1, 2013 at 10:22am by Kim*

**Complex Number Proof**

Prove: If Z and W are complex numbers, then the conjugate of (Z+W) is equal to the conjugate of Z plus the conjugate of W. My thought is that this is kind of like the distributive property, but I'm not sure. It doesn't help that I haven't written a proof in over 10 years. Help...
*Wednesday, June 11, 2008 at 4:38pm by Amy*

**algebra**

Can someone please help me with these problems? Please and thank you. 1)Solve the equation: (1/x)+[x/(x+2)]=1 2)Simplify the complex fraction: 3-(3/4) ------- (1/2)-(1/4) (1/x)+ [x/(x+2)] = 1 Convert left side to the sum of fractional terms with common denominators. [(x+2) + x...
*Sunday, May 13, 2007 at 11:28am by Lydia*

**Complex Numbers**

You are welcome
*Monday, January 31, 2011 at 9:03pm by helper*

**math**

in real numbers there is no solution, since x^2 is always positive, so adding 9 will never be zero. In complex numbers, the solutions are 3i and -3i, since i^2 = -1.
*Friday, November 23, 2012 at 2:41am by Steve*

**help**

My friend, area of complex figure is complex to find without given constraints. Ok, maybe you divide complex shape into familiar shapes. Example: complex figure is made from square and two triangles. Then is easy. So divide complex figure into little pieces of easy figures. ...
*Thursday, March 7, 2013 at 5:26pm by sputnik*

**LAL!**

are these complex sentence. if it is complex, and are the subordinate clause right are these right? the subordinate clause are in capital letters 1-IN THE 1950S the giants were the national league team from new york.( COMPLEX) 2- fans argued all the time about WHO WAS THE ...
*Wednesday, May 12, 2010 at 3:56pm by Sara!*

**english/check please**

Find the independent clauses and the dependent clauses.Write wether it is complex or compound-complex 1)Mr and Mrs Johnson play golf(ind) when the weather is warm.(dep)complex 2)What I would like is a vacation. independent,complex 3)Alicia works at the daycare center that is ...
*Monday, April 30, 2012 at 11:37pm by joe*

**Calculus--complex numbers**

thanks!
*Thursday, March 13, 2008 at 7:11am by Anne---?*

**complex numbers**

Thank you so much!!!
*Tuesday, November 4, 2008 at 1:43pm by Mary*

**Trigonometry**

complex numbers
*Sunday, December 8, 2013 at 10:57pm by steve*

**english**

Write whether a sentence is simple, compound, complex, or compound-complex. 1-Mrs. Huang planned a party for Kim because she was graduating from high school. Complex 2=She invited Lisa,Elena,and Enrique to the party. Simple 3-Mr. Huang thought that kim's friends should come to...
*Tuesday, May 1, 2012 at 9:28pm by joe*

**English**

When you are looking at a work by Monet, stand back at least fifteen or twenty feet. simple, compound, complex, compound-complex complex
*Wednesday, November 17, 2010 at 9:58pm by Hayley*

**English**

1)compound 2)complex 3)complex 4)compound complex 5)simple 6)simple 7)simple 8)simple 9)compound 10)compound 11)complex 12)complex 13)simple
*Tuesday, November 9, 2010 at 2:10pm by Tom Jameson*

**chemistry**

How to find moles given oxidation numbers and chemical formula? Trying to do my chemistry lab but i'm super duper stuck. please help me i'm gonna die. this is the part i'm stuck on: The overall charge on the complex, [Ni(en)(H2O)]^2+ * SO4^2- * H2O is zero; the charge on ...
*Thursday, November 15, 2012 at 12:37am by dr bob's number one fan*

**Algebra 2**

When dealing with imaginary and complex numbers, are there 2 solutions? I.e.: x^2=4 x=2,-2 For: x^2=-4 x=2i or -2i ? (Is there a 2i and -2i, like for real numbers?)
*Wednesday, October 24, 2012 at 1:00pm by Bart*

**math-complex numbers**

all are circles of radius 2, with center at various numbers. |z-c| = r has center at c. |z|^2 = x^2+y^2 That's why the curves are circles.
*Wednesday, October 9, 2013 at 9:34pm by Steve*

**maths-complex numbers**

by using the substitution w = z^3, find all the solutions to z^6 - 8z^3 +25 = 0 in complex numbers, and describe them in polar form, using @(theta) to denote the angle satisfying tan@ = 3/4 ( note simply leave @ as it is, dont calculate it). i got up to z^3 = 4+3i and 4-3i ...
*Monday, March 12, 2007 at 9:25am by Anonymous*

**Calculus--complex numbers**

thank you so much!!
*Wednesday, March 12, 2008 at 4:09am by Anne*

**complex numbers**

if 5i=yi, then y is 5 and if y is 5, x=y
*Sunday, March 8, 2009 at 2:34pm by bobpursley*

**Math**

These are complex numbers by the way.
*Monday, November 23, 2009 at 10:35am by Nayla*

**Math (Complex Numbers)**

76
*Thursday, May 9, 2013 at 9:55am by Athul*

**English**

As it happens, you're right and I'm wrong. Tell if compound, complex, simple, compound-complex complex
*Wednesday, November 17, 2010 at 9:45pm by Hayley*

**Algebra 2/Trig.**

I forgot some stuff from my 7th grade algebra class about complex numbers...like what do you do with a problem like this 5+5i 3-i divided by + divided by 2-i 4+3i thanks.(divided by refers to dividing line in a fraction) In each case, RATIONALIZE the denominator. Here is the ...
*Sunday, November 19, 2006 at 7:08pm by Tom*

**Sociology**

6. President Eisenhower was a) A supporter of the military industrial complex b) The founder of the military industrial complex c) A general so he must be in favor of the military industrial complex d) Concerned that the military industrial complex would grow too big and ...
*Sunday, September 16, 2012 at 7:38pm by Tammy*

**math b**

divide complex numbers: 2+3-3i^3/1-i
*Wednesday, April 1, 2009 at 8:48pm by annaiiz*

**Math**

simplify (8 – i) – (4 – i) using Complex Numbers
*Wednesday, September 7, 2011 at 12:43am by Peter*

**algebra**

I would use the standard quadratic formula for ax²+bx+c=0, and x=(-b±√(b²-4ac))/2a Here 4x^2 - 5x = -2 can be rewritten as 4x^2 - 5x + 2 =0, so a=4, b=-5 c=2 substituting in the quadratic formula, x=(-(-5)±√((-5)²-4(4)(2)))/(2*4) which ...
*Tuesday, June 8, 2010 at 8:42am by MathMate*

**Maths**

almost sinx = cos(pi/2 - x) and cosx = sin(pi/2 - x) that is, the sine of any angle equals the cosine of its compliment. so sinx - i cosx = cos(pi/2 - x) - i sin(pi/2 - x) = cis(pi/2 - x) this webpage has some good examples with simple diagrams to show your topic http://www....
*Wednesday, March 4, 2009 at 2:44am by Reiny*

**complex numbers**

find the values of x and y x+5i=y+yi
*Sunday, March 8, 2009 at 2:34pm by Lou*

**algebra**

x^2+225 ..factor over complex numbers?
*Tuesday, April 6, 2010 at 11:57pm by mada*

**Algebra 2**

For any polynomial of odd degree, large negative values of x yield large negative values for y. Same for positive values. So, somewhere along the way, y must change from negative to positive, meaning there is at least one real root. Now, it's quite possible that there is only ...
*Saturday, May 18, 2013 at 9:22am by Steve*

**Math**

Michele's answer is not correct x^=-9 has no solution in the set of real numbers (x=-3 is a real number solution) You will have to use imaginary or complex numbers x^2 = -9 x = ±√-9 x = ± 3i where i = √-1
*Monday, October 8, 2007 at 12:44pm by Reiny*

**math**

2y=7sqrt(y)-3 2y+3 = √y square both sides 4y^2 + 12y + 9 = y 4y^2 + 11y + 9 = 0 y = (-11 ±√-23)/8 I am assuming you are dealing with real numbers only since y is a complex number, and we cannot take a square root of a complex number, your equation has no real ...
*Friday, July 26, 2013 at 10:11pm by Reiny*

**Calculus**

two complex numbers Z and W are related by the formula W=(Z+1)/(1-Z) what is W if Z=1.5i? thanks for the help!
*Thursday, March 13, 2008 at 7:15am by Anne*

**math**

no, i entered it correctly. the unit is complex numbers and polynomials
*Tuesday, February 24, 2009 at 8:31pm by isha*

**math b**

last question finally simplify complex numbers: i^16/i^3
*Wednesday, April 1, 2009 at 9:45pm by annaiiz*

**algebra**

x^2+225 ... factor expression over the complex numbers?
*Wednesday, April 7, 2010 at 12:00am by mada*

**math**

Simplify the following product of complex numbers. (8 + i) • (-4 – 3i)
*Wednesday, March 2, 2011 at 9:03pm by Jackster*

**Math**

Factor over {complex numbers} 9x^2 + 49
*Monday, April 9, 2012 at 9:34am by Nathan*

**Math**

using only real numbers there is no factorization. Using complex numbers, 16x^2 + 49y^2 = (4x+7y i)(4x-7y i)
*Tuesday, November 6, 2012 at 11:02pm by Steve*

**Math**

Use the quadratic formula to solve the equation. x² - x = -2 Type an exact answer, using radicals as needed. Express complex numbers in terms of i. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
*Sunday, April 24, 2011 at 10:07am by Mary*

**college algebra--need help please!!**

From the given zeros, factors would be (x+6) , (x-6-i) and (x-6 + i) the complex factors simplify to x^2 -12x +37 (complex numbers always appear in conjugate pair form) but that would only produce a cubic, so there has to be multiplicity of the above zeros. two options : f(x...
*Friday, November 23, 2012 at 2:59pm by Reiny*

**math complex numbers**

oops, I forgot the 3i, so the answer is 4+i check that.
*Wednesday, November 17, 2010 at 8:45pm by bobpursley*

**Math - complex numbers**

Sorry, yes I meant from 0 to pi/2 for the last two questions.
*Saturday, December 17, 2011 at 12:59am by Chiquesta*

**Math**

I Don't Know What Is This Subject Called In English But It's About Complex Numbers it says 25 = 16 - 9 i^2 = (4 ± 3i) But My Teacher Says 25 = 25 - 0 i^2 = (5 ± 0i) I Think My Teacher Is Wrong , The Result Is 25 But There Will Not Be Complex Number Because ( 0i = 0) So Is My ...
*Wednesday, October 5, 2011 at 6:59am by Jack*

**English**

Please check my work. Identify whether each sentence is complex, an exclamation, declarative, or interrogative. 1. Tim likes soup, so Mary bought tomatoes. Complex 2. Oh no! Exclamation 3. Do you want to play in the Garden? Interrogative 4. In the story, the author told how ...
*Wednesday, July 20, 2011 at 12:32am by Priya*

**English**

Please check my work. Identify whether each sentence is complex, an exclamation, declarative, or interrogative. 1. Tim likes soup, so Mary bought tomatoes. ----- Complex 2. Oh no! ----- Exclamation 3. Do you want to play in the Garden? ----- Interrogative 4. In the story, the ...
*Wednesday, July 20, 2011 at 9:34am by Priya*

**math URGENT!!!!**

i have a list of 5 different numbers. the sum of 2 of the numbers is 0. the sum of 2 of the numbers is 1. the sum of 2 of the numbers is 2. the sum of 2 of the numbers is 3. the sum of 2 of the numbers is 4.the sum of 2 of the numbers is 5.the sum of 2 of the numbers is 6. the...
*Monday, October 29, 2012 at 5:14pm by dan*

**intermediate algebra**

Simplify the following product of complex numbers (-10+3i)*(-2)
*Saturday, July 9, 2011 at 9:48pm by swirl*

**maths (complex numbers)**

if b is a real number satisfying b^4 + 1/(b^4) = 6, find the value of :- (b + i/b)^16 where i = sqrt of -1
*Wednesday, October 26, 2011 at 2:00am by ...........*

**maths**

Find the number of different ordered quadruples (a,b,c,d) of complex numbers such that a^2=1 b^3=1 c^4=1 d^6=1 a+b+c+d=0
*Monday, April 1, 2013 at 10:22am by stranger*

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