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Normal CO2 in the blood is the same as Normal PH.
Ph is 7.35-7.45
CO2 is 35-45
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How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
What is the integral of sine to the 7th power 3x dx?
If you look in most calculus books there is a table of intergals. Your problem is one of the trigonometric forms. Below is given the reduction formula,∫sinn(u) du = (-1/n)sinn-1(u)cos(u) + ((n-1)/n)∫sinn-2(u) duwhere n ≥ 2 is an integer.Basically yours would be n=7 & u=3x. You keep integrating until you lose the last integral sign on the right side. Pretty tedious work, but if I am using the right integral form & didn't make any mistakes the answer should be:(-1/7)sin6(3x)cos(3x) - (6/35)sin4(3x)cos(3x) - (8/35)sin2(3x)cos(3x) - (16/35)cos(3x) + CHere is the work:Let u = 3xu' = 3 dx, so that dx = du/3∫sin7 3x dx = (1/3)∫sin7(u) du∫sin7(u) du = (-1/7)sin6(u)cos(u) + (6/7)∫sin5(u) du= (-1/7)sin6(u)cos(u) + (6/7)[(-1/5)sin4(u)cos(u) + (4/5)∫sin3(u) du]= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)∫sin3(u) du= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)[(-1/3)sin2(u)cos(u) + (2/3)∫sin(u) du]= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) + (16/35)∫sin(u) du= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u) + Cso that∫sin7 3x dx = (1/3)∫sin7(u) du= (1/3 [(-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u)] + C= -(1/21)sin6(u)cos(u) - (2/35)sin4(u)cos(u) - (8/105)sin2(u)cos(u) - (16/105)cos(u) + C= -(1/21)sin6(3x)cos(3x) - (2/35)sin4(3x)cos(3x) - (8/105)sin2(3x)cos(3x) - (16/105)cos(3x) + COr∫sin7 3x dx = ∫sin63x sin 3x dx = ∫(sin2 3x)3 sin 3x dx =∫(1 - cos2 3x)3 sin 3x dxLet u = cos 3xu' = (cos 3x)'du = -sin 3x*3 dx, and dx = du/-3sin 3x= (-1/3)∫(1 - u2)3 du = ∫(1 - 3u2 + 3u4 + u6) du = (-1/3) [u - (3/3)u3 + (3/5)u5 - u7/7] + C= (-1/3)u + (1/3)u3 - (1/5)u5 + (1/21)u7 + C= (-1/3)cos 3x + (1/3)cos3 3x - (1/5)cos5 3x + (1/21)cos7 3x + C
In a right triangle sinA 35 What is cos A?
Sin A must be a number whose absolute value cannot exceed 1 and so it cannot be 35.
How do you find the chord of a 35 degrees arc in a circle of radius 14 inches?
Use the Law of Cosines: c^2 = a^2+b^2-2ab cos C. Here, the two legs of a triangle are both the radius (14 inches long), and the angle between them is 35 degrees. So c^2 = 14^2+14^2-2*14*14*cos 35 c^2 = 2*14^2(1-cos 35) c^2 = 70.89.. c = 8.42..
How will the value of sin 35 determine?
to find sin 35 here we take the angle = x=15 then 3x=45 , 4x=60 then 4x-3x=60-45 then by putting sin on rhs we will get cos 35 and sin 35 hope it helped you
What is the numerical value of cos 35 x sin 24?
In degrees? cos(35˚) = .81915, sin(24˚) = .40673;cos(35˚) * sin(24˚) = .33318In radians? cos(35) = -.90367, sin(24) = -.90558;cos(35) * sin(24) = .81836A calculator will achieve these results faster than wiki.answers. 9 times out of 10, at least.:-)
Cosine 35 degrees sine 55 degrees plus sine 35 degrees cosine 55 degreees?
cos(35)sin(55)+sin(35)cos(55) If we rewrite this switching the first and second terms we get: sin(35)cos(55)+cos(35)sin(55) which is a more common form of the sin sum and difference formulas. Thus this is equal to sin(90) and sin(90)=1
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
What is the integral of sine to the 7th power 3x dx?
If you look in most calculus books there is a table of intergals. Your problem is one of the trigonometric forms. Below is given the reduction formula,∫sinn(u) du = (-1/n)sinn-1(u)cos(u) + ((n-1)/n)∫sinn-2(u) duwhere n ≥ 2 is an integer.Basically yours would be n=7 & u=3x. You keep integrating until you lose the last integral sign on the right side. Pretty tedious work, but if I am using the right integral form & didn't make any mistakes the answer should be:(-1/7)sin6(3x)cos(3x) - (6/35)sin4(3x)cos(3x) - (8/35)sin2(3x)cos(3x) - (16/35)cos(3x) + CHere is the work:Let u = 3xu' = 3 dx, so that dx = du/3∫sin7 3x dx = (1/3)∫sin7(u) du∫sin7(u) du = (-1/7)sin6(u)cos(u) + (6/7)∫sin5(u) du= (-1/7)sin6(u)cos(u) + (6/7)[(-1/5)sin4(u)cos(u) + (4/5)∫sin3(u) du]= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)∫sin3(u) du= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)[(-1/3)sin2(u)cos(u) + (2/3)∫sin(u) du]= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) + (16/35)∫sin(u) du= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u) + Cso that∫sin7 3x dx = (1/3)∫sin7(u) du= (1/3 [(-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u)] + C= -(1/21)sin6(u)cos(u) - (2/35)sin4(u)cos(u) - (8/105)sin2(u)cos(u) - (16/105)cos(u) + C= -(1/21)sin6(3x)cos(3x) - (2/35)sin4(3x)cos(3x) - (8/105)sin2(3x)cos(3x) - (16/105)cos(3x) + COr∫sin7 3x dx = ∫sin63x sin 3x dx = ∫(sin2 3x)3 sin 3x dx =∫(1 - cos2 3x)3 sin 3x dxLet u = cos 3xu' = (cos 3x)'du = -sin 3x*3 dx, and dx = du/-3sin 3x= (-1/3)∫(1 - u2)3 du = ∫(1 - 3u2 + 3u4 + u6) du = (-1/3) [u - (3/3)u3 + (3/5)u5 - u7/7] + C= (-1/3)u + (1/3)u3 - (1/5)u5 + (1/21)u7 + C= (-1/3)cos 3x + (1/3)cos3 3x - (1/5)cos5 3x + (1/21)cos7 3x + C
Is 35 degrees cos or sin?
Every angle has a sine and a cosine. The sine of 35 degrees is 0.57358 (rounded) The cosine of 35 degrees is 0.81915 (rounded)
If f(x) cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then find the limit of 1 - f(x)35(sinx)2 as x tends to 0 (zero).?
I'm sorry the question is not correctly displayed. If f(x) = cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then, find the limit of {1 - [f(x)]^3}/[5(sinx)^2] as x tends to 0 (zero).
In a right triangle sinA 35 What is cos A?
Sin A must be a number whose absolute value cannot exceed 1 and so it cannot be 35.
How do you find the chord of a 35 degrees arc in a circle of radius 14 inches?
Use the Law of Cosines: c^2 = a^2+b^2-2ab cos C. Here, the two legs of a triangle are both the radius (14 inches long), and the angle between them is 35 degrees. So c^2 = 14^2+14^2-2*14*14*cos 35 c^2 = 2*14^2(1-cos 35) c^2 = 70.89.. c = 8.42..
How will the value of sin 35 determine?
to find sin 35 here we take the angle = x=15 then 3x=45 , 4x=60 then 4x-3x=60-45 then by putting sin on rhs we will get cos 35 and sin 35 hope it helped you
What is cos square?
Cos times Cos
Is cos squared x the same as cos x squared?
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
Cos plus cos plus cos?
3cos
