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####Nadine####
Questions (5)
When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 8.80 cm wide on a
7 answers
4,005 views
A 2.11 cm high insect is 1.31 m from a -125 mm focal-length lens. Where is the image?
Distance of the image is ? How high is it?
3 answers
658 views
When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 8.80 cm wide on a
1 answer
730 views
A 2.11 cm high insect is 1.31 m from a 125 mm focal-length lens. Where is the image?
Distance of the image is ? How high is it?
0 answers
1,034 views
An aquarium filled with water has flat glass sides whose index of refraction is 1.48. A beam of light from outside the aquarium
2 answers
2,817 views
Answers (7)
x1
Watch your signs! (x^2 -4x+4)-4(y^2 +6y+9)=36+4+36 should be (x^2 -4x+4)-4(y^2 -6y+9)=36+4-36 (x^2 -4x+4)/4-4(y^2 -6y+9)/4=4/4 ((x-2)^2)/4- (y-3)^2 =1 hyperbola, center at (2,3)
Thank you so much Elena! I forgot to divide 8.8 by two and, of course, kept getting double the answer which I knew was incorrect.
Thank you Elena, I had virtual and inverted also, but the book says it is virtual and upright. so I thought everything else was worry. Maybe it is the book?
Chain and quotient rule ((1/2)(2x)(x^2-1)^-1/2)x^2 -2x(x^2-1)^1/2)/x^4 simplifies to (x^3(x^2-1)^-1/2 -2x(x^2-1)^1/2)/x^4 simplifies to (x^2-1)^-1/2(x^2)-2(x^2-1))/x^3 final answer ((x^2-1)^(-1/2))(-x^2+2)/x^3
1. .5 = e^1.5k ln .5 = 1.5 k ln.5÷1.5 = k m(4) = 400e^4(ln.5÷1.5)=62.996 mg m(8) = 400e^8(ln.5÷1.5)=9.921 mg 2. If half life =t1/2 .5 = e^kt1/2 ln.5 = kt1/2 k=ln.5/t1/2 hr 3. .01 = e^t(ln.5÷1.5) ln.01 =t( ln.5 ÷1.5) ln .01/( ln.5 ÷1.5)=t 9.966 hr=t
Thank you, I did that but I did not round until the end so I had 31.06º which was listed as incorrect
