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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
What else can I help you with?
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..
A picture of weight 25N hangs from a hook by two strings which each makes an angle of 35 with the vertical .What is the tension in each string?
To find the tension in each string, we can use the concept of equilibrium. The vertical components of the tension in both strings must equal the weight of the picture (25 N). Since the angle is 35 degrees, the vertical component of the tension ( T ) in each string is ( T \cos(35^\circ) ). Therefore, the equation becomes ( 2T \cos(35^\circ) = 25 ). Solving for ( T ) gives ( T = \frac{25}{2 \cos(35^\circ)} \approx 21.5 , \text{N} ).
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.:-)
What is the value of tan of 35 degrees?
Tan(35) = 0.700207.... However, Tan = Sin/ Cos Hence Tan(35) = Sin(35) / Cos(35) = 0.57357... / 0.81915... & 0.57357... / 0.81915.... = 0.7002.... As before!!!!!
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..
A picture of weight 25N hangs from a hook by two strings which each makes an angle of 35 with the vertical .What is the tension in each string?
To find the tension in each string, we can use the concept of equilibrium. The vertical components of the tension in both strings must equal the weight of the picture (25 N). Since the angle is 35 degrees, the vertical component of the tension ( T ) in each string is ( T \cos(35^\circ) ). Therefore, the equation becomes ( 2T \cos(35^\circ) = 25 ). Solving for ( T ) gives ( T = \frac{25}{2 \cos(35^\circ)} \approx 21.5 , \text{N} ).
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
