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maths
Hi, could someone please tell me how I would go about working these out: (the answer must be in the form a(sqroot)b where a and b are integers and b is as small as possible.) 1) Simplify (sqroot)500 2) Simplify (sqroot)50 x
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Simplify 3(x2)^2/x2 Is the answer 3x^210x+10? If not then what is? When p= sqrt2 and q= sqrt6 write pq in the form a(sqrt)b where a and b are integers and b is as small as possible. Multiply out these brackets. Give your answer
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ALGEBRA 1
change 50 to 25*2 sqrt 50 = 5 sqrt2 change 8 to 4*2 Nope: (sqrt3sqrt5)(sqrt3sqrt5). remember (ab)(ab)= a^22ab + b^2 Simplify.Express in simpliest radical form. 1.3sqrt2 +sqrt50 Cannot be solve. Terms have unlike radicals
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evaluate and simplify 6  square root20  2 The square root of 20= sqrt (4*5)= sqrt4 * sqrt5= 2 sqrt5 The square root of 20= sqrt (4*5)= sqrt4 * sqrt5= 2 sqrt5
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algebra,math,help
Use Property 2 to simplify each of the following radical expressions. sqrt (10)/ sqrt(49) My answer: sqrt (10) / (7) THis next one i need help: Use the properties for radicals to simplify each of the following expressions. Assume
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Can someone help me with this problem please? rationalize the denominator and simlify the answer: 5/2 + the square root of 3 Do you mean 5/(2 + sqrt 3) ? What you wrote could also be interpreted as 2.5 + sqrt 3. You need to be
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Math  check my answers :)
1. Find the amplitude, if it exists, and the period of y=tan 1/4È Answer: No limits in amplitude, y=tanÈ/4 period is 4pi 2.Solve x = Arctan(sqrt3) Work: Pythagorean theorem: sqrt(1^2+3)= sqrt4=2 in length. Triangle is 1, sqrt
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Two more I would like to check my work on: Solve using the square root property: (x+6)^2 = 4 x+6 = sqrt4 x+6 = sqrt4 x+6 = 2 x+6 = 2 x+66 = 26 x+66 = 26 x = 4 x = 8 Solve by completing the square: x^2+6x+1=0 x^2+6x+11 =
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Algebra
Multiplying sq rts sqrt18a^7b times sqrt27a^8b^6 Jake 1214 18 = 9 * 2 then sqrt 18 = 3sqrt2 sqrt a^7 = a^3 * sqrt a 27 = 9 * 3 then sqrt 27 = 3 sqrt3 sqrt a^8 = a^4 sqrt b^6 = b^3 Now just multiply the liketerms together. 3sqrt6
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Express log A in terms of the 2 logarithms of prime numbers: A= (sqrt(3)*sqrt4(125))/7^3
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