Tutors Answer Your Questions about Radicals (FREE)
Question 1196120: The frequency of a wave is the number of oscillations it completes per unit of time Frequency is measured in oscillations per second One second is equal to one hertzThe frequency of an object attached to a spring can be calculated using the formulacycles per secondIn this formula, k is a constant that depends on the material that is and stands for the mass of the object Suppose you place a 4-kg weight at the end of a spring Complete parts (a) and (b) f = 1/(2pi) * sqrt(k/m)
Click here to see answer by ikleyn(52776)   |
Question 1196120: The frequency of a wave is the number of oscillations it completes per unit of time Frequency is measured in oscillations per second One second is equal to one hertzThe frequency of an object attached to a spring can be calculated using the formulacycles per secondIn this formula, k is a constant that depends on the material that is and stands for the mass of the object Suppose you place a 4-kg weight at the end of a spring Complete parts (a) and (b) f = 1/(2pi) * sqrt(k/m)
Click here to see answer by Alan3354(69443)  |
Question 1196335: Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
Click here to see answer by josgarithmetic(39616) |
Question 1196335: Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
Click here to see answer by ikleyn(52776) |
Question 1196337: [ I am reposting because there was a typo in my email when I posted the previous question ]
Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
Click here to see answer by josgarithmetic(39616) |
Question 1196337: [ I am reposting because there was a typo in my email when I posted the previous question ]
Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
Click here to see answer by math_tutor2020(3816) |
Question 1196885: v
=
√
20
L
to estimate the speed of a car,
v
, in miles per hour, based on the length,
L
, in feet, of its skid marks when suddenly braking on a dry, asphalt road. At the scene of an accident, a police officer measures a car's skid marks to be 163 feet long. Approximately how fast was the car traveling? Round your answer to the nearest tenth (one decimal place) of a unit.
Click here to see answer by Alan3354(69443) |
Question 1196885: v
=
√
20
L
to estimate the speed of a car,
v
, in miles per hour, based on the length,
L
, in feet, of its skid marks when suddenly braking on a dry, asphalt road. At the scene of an accident, a police officer measures a car's skid marks to be 163 feet long. Approximately how fast was the car traveling? Round your answer to the nearest tenth (one decimal place) of a unit.
Click here to see answer by ikleyn(52776) |
Question 1199216: Solve for x.
x^(3/2) - 3x^(1/2) = 0
sqrt{x^2} - sqrt{3x} = 0
After squaring both sides, I got x^3 = 3x.
x^3 - 3x = 0
x(x^2 - 3) = 0
x = 0 and x^2 - 3 = 0
Solving x^2 - 3 = 0 for x, I got sqrt|3}, -sqrt{3}.
After checking my answers for x, I concludes that the only two solutions
for x are 0 and sqrt{3}.
The textbook answers for x are 0 and 3.
Why am I wrong?
Click here to see answer by ikleyn(52776) |
Question 1199216: Solve for x.
x^(3/2) - 3x^(1/2) = 0
sqrt{x^2} - sqrt{3x} = 0
After squaring both sides, I got x^3 = 3x.
x^3 - 3x = 0
x(x^2 - 3) = 0
x = 0 and x^2 - 3 = 0
Solving x^2 - 3 = 0 for x, I got sqrt|3}, -sqrt{3}.
After checking my answers for x, I concludes that the only two solutions
for x are 0 and sqrt{3}.
The textbook answers for x are 0 and 3.
Why am I wrong?
Click here to see answer by greenestamps(13198)  |
Question 1199216: Solve for x.
x^(3/2) - 3x^(1/2) = 0
sqrt{x^2} - sqrt{3x} = 0
After squaring both sides, I got x^3 = 3x.
x^3 - 3x = 0
x(x^2 - 3) = 0
x = 0 and x^2 - 3 = 0
Solving x^2 - 3 = 0 for x, I got sqrt|3}, -sqrt{3}.
After checking my answers for x, I concludes that the only two solutions
for x are 0 and sqrt{3}.
The textbook answers for x are 0 and 3.
Why am I wrong?
Click here to see answer by math_tutor2020(3816) |
Question 1201554: The velocity,v, in meters per second, of a roller coaster at the bottom of the hill is related to the vertical drop, h, in meters, and the velocity, v0, in meters per second, of the roller coaster at the top of the hill by the formula v0 = √v²-20h.
a.) Tom simplifies the expression for the formula to v0 = v - 2 √5h. is he correct. explain you reasoning.
b.) Suppose the velocity at the top of the hill is 20m/s and the velocity at the bottom of the hill is 40m/s. What is the vertical drop of the hill
Click here to see answer by mananth(16946)  |
Question 1206414: Question: sqrt(x)-sqrt(3x-3)=1
My work:
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.
Click here to see answer by math_tutor2020(3816) |
Question 1206414: Question: sqrt(x)-sqrt(3x-3)=1
My work:
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.
Click here to see answer by ikleyn(52776) |
Question 1206414: Question: sqrt(x)-sqrt(3x-3)=1
My work:
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.
Click here to see answer by MathTherapy(10551)  |
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