a^3=1/cos-cos = (1-cos^2)/cos=sin^2/cos b^3=1/sin-sin = (1-sin^2)/sin=cos^2/sin a^3/b^3=sin^2/cos * cos^2/sin=sin^3/cos^3 => a/b=sin/cos = tan (Hope you understand your own notations.)
μ=98.2 σ=0.62 Z(100)=(100-98.2)/0.62=2.903 Look up (normal) probabilities for Z=2.903 from tables (for people with temperatures below 100). Subtract from 1 to get probabilities of healthy person considered having a fever.
Assuming radiator reaches 120° between 0 and 4 minutes. Trying to solve: 120 = 6.2t² + 12t +32 or 6.2t² + 12t - 88 = 0 Use quadratic formula to get t=-4.8 or t=2.92 sec. Reject t=-4.8 because it is not within interval [0,4].
1. The transformations are: e(O,2) : dilation about origin of 2 units t(0,5) : translation 5 units up r(y,-x) : rotation counter-clockwise 90° So (x,y)-> (2x,2y) -> (2x, 2y+5) -> (2y+5, -2x) So combining it all: (x,y) -> (2y+5, -2x) 2. Using Cramer's rule: ...
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Math 4 questions
"at most" means ≤ "minimuum of" means ≥ "less than" means < "low of" means ≥ "high of" means ≤ You will be more likely to get an answer quickly if you show your attempts.
Algebra II Part 2
Thanks Steve, good thought!
Algebra II Part 2
use the identities: tan(x)=sin(x)/cos(x) cot(x)=cos(x)/sin(x) and sin(π/2)=1, cos(π/2)=0 sin(0)=0, cos(0)=1
Use compound interest formula: Amount = Principal * (rate)^n rate=interest rate per period n=number of periods compounded. Here rate = 1+ 0.07/2=1.035 number of periods = 3 years * 2 periods/yr principal=$7300 The question didn't say what's required.
Let the mathematical model be f(x)=ax^b where x=years past 1970. f(x)=population in millions Note: much better accuracy is obtained by shifting the origin to 1970, the closest possible. Substituting given data for 1970: 5.1=ax^0=a so f(x)=5.1x^b For 2004, 14.3=5.1(34^(b)) or ...
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