Pre calc
cosx/1+sinx or cosx/(1+sinx) 1+six/cosx or (1+sinx)/cosx ?
Trig
Multiply both sides by sin^2(x). sin^2(x)=cos^2(x)+sin(x) sin^2(x)=1-sin^2(x)+sin(x) 2sin^2(x)-sin(x)-1=0 (2sin(x)+1)(sin(x)-1)=0 sin(x)=-1/2, x=7pi/6, x=11pi/6 or sin(x)=1, x=pi/2
math
342
math 12
I think that nobody knows how to solve it algebraically. If x=16 left side=0, right side=2.3*10^(-10) Ask your teacher again.
math
Multiply both sides by log16. log16/log2=4 log16/log4=2 log16/log8=4/3 log16/log16=1 logx(4+2+4/3+1)=25*log16 logx=3*log16 logx=log(16^3) x=16^3=4096 (check the answer at the back)
algebra
(-2)^5
Math - Possible or Potential rational zeros
Potential rational zeros are: +-1,+-2,+-7,+-14 f(-1)=1+7-3-19+14=0 f(x)=(x+1)g(x), g(x)=x^3-8x^2+5x+14 g(-1)=-1-8-5+14=0 g(x)=(x+1)h(x), f(x)=(x+1)^2*h(x), h(x)=x^2-9x+14 h(2)=4-18+14=0 h(x)=(x-2)(x-7) f(x)=(x-2)(x-7)(x+1)^2
Math - Real number system
Let f(x)=x^4+6x^3+x^2-24x-20 Real roots are looking for among +-1,+-2,+-4,+-5,+-10,+-20 f(-1)=1-6+1+24-20=0 Divide f(x) by (x+1): f(x)=(x+1)(x^3+5x^2-4x-20) Let g(x)=x^3+5x^2-4x-20 g(-5)=-125+125+20-20=0 Divide g(x) by (x+5): g(x)=(x+5)(x^2-4) f(x)=(x+1)(x+5)(x+2)(x-2)
Math
If the function f(x) is continuous on [a,b] and f(a)*f(b)<0 then on (a,b) there are at least one zero. Polynomial is continuous on R. Compute f(-2.8) and f(-2.7)
Math
i^3=-i s^2i ???
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