math
you know the values for 30 and 45 degrees use your sum/difference and half-angle formulas 75=45+30 15=30/2 105=90+15
Math (Algebra)
If the coefficients are all rational, then the irrational roots must occur in conjugate pairs. So, the polynomial is (x-(√2+√3+√5)) (x-(√2+√3-√5)) (x-(√2-√3+√5)) (x-(√2-√3-√5)) (x-(-√2+√3+ͩ...
algebra
g+(4g+8)=30 5g=22 No integer solution. Is there a typo? checking, g=5 b=28 makes 33 g=4 b=24 makes 28 No way to make 30 total
algebra
14*7 x^4 x^3 = 98x^7
Precalc Math
5000=P(1+.035/12)^(12*4) P = 4347.68
calculus
v=pi r^2 h = 25, so h = 25/(pi r^2) a = 2πr^2 + 2πrh = 2πr^2 + 2πr*25/(πr^2) = 2πr^2 + 50/r da/dr = 4πr - 50/r^2 = (4πr^3 - 50)/r^2 since r≠0, da/dr=0 when 4πr^3 = 50 r = ∛(25 / 2π)
Calculus
Assuming the usual carelessness with parentheses, I gather you mean f(x) = ((x-3)/(2x+3))^2 If f = u^2, f' = 2u u' So, f' = 2(x-3)/(2x+3) * ((x-3)/(2x+3))' = 2(x-3)/(2x+3) * [(2x+3)-2(x-3)/(2x+3)^2] = 2(x-3)/(2x+3)*(9/(2x+3)^2) = 18(x-3)/(2x+3)^3
math
128y = -4x^2 - 24x + 988 32y = -x^2 - 6x + 247 32y = -(x^2+6x+9) + 247+9 32y = -(x+3)^2 + 256 y = -1/32 (x+3)^2 + 8
algebra
a^1/2 b * 1/2 (a b^1/2) 1/2 a^3/2 b^3/2
pre algebra
only the center cube, so 1
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