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Question Number 152692 Answers: 0 Comments: 0

If b and h are two integers with b>h, and b^2 +h^2 =b(a+h)+ah, find the value of b.

$$\mathrm{If}\:{b}\:\mathrm{and}\:{h}\:\mathrm{are}\:\mathrm{two}\:\mathrm{integers}\:\mathrm{with}\:{b}>{h}, \\ $$$$\mathrm{and}\:{b}^{\mathrm{2}} +{h}^{\mathrm{2}} ={b}\left({a}+{h}\right)+{ah}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{b}. \\ $$

Question Number 152691 Answers: 1 Comments: 0

If a,p,q are primes with a<p, and a+p=q, find the value of a.

$$\mathrm{If}\:{a},{p},{q}\:\mathrm{are}\:\mathrm{primes}\:\mathrm{with}\:{a}<{p},\:\mathrm{and} \\ $$$${a}+{p}={q},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}. \\ $$

Question Number 152683 Answers: 1 Comments: 0

The probability that athlete will win a race is (1/6) and that he will be second and third are (1/4) and (1/3) respectively.what is the probability that he will not be first in the first three place! Please,help me out

$${The}\:{probability}\:{that}\:{athlete}\:{will}\:{win}\:{a}\:{race}\:{is}\:\frac{\mathrm{1}}{\mathrm{6}}\:{and}\:{that} \\ $$$${he}\:{will}\:{be}\:{second}\:{and}\:{third}\:{are}\:\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${respectively}.{what}\:{is}\:{the}\:{probability}\:{that}\:{he}\:{will}\:{not}\:{be}\:{first} \\ $$$${in}\:{the}\:{first}\:{three}\:{place}! \\ $$$${Please},{help}\:{me}\:{out} \\ $$

Question Number 152682 Answers: 1 Comments: 0

Question Number 152672 Answers: 1 Comments: 0

Question Number 152671 Answers: 3 Comments: 2

Question Number 152670 Answers: 1 Comments: 0

Question Number 152678 Answers: 2 Comments: 0

Find the sum of the real roots of equations: a^3 -6a^2 +15a-17=0 and a^3 -6a^2 +15a-11=0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{17}=\mathrm{0}\:\:\:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{11}=\mathrm{0} \\ $$

Question Number 152663 Answers: 1 Comments: 0

If x^3 -x+3=0 has the roots a, b and c. determine the monic polynomial with the roots a^5 , b^5 and c^5 . [Q152396]

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} -\mathrm{x}+\mathrm{3}=\mathrm{0}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{monic}\:\mathrm{polynomial}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\:\mathrm{a}^{\mathrm{5}} ,\:\mathrm{b}^{\mathrm{5}} \:\mathrm{and}\:\:\mathrm{c}^{\mathrm{5}} . \\ $$$$\left[{Q}\mathrm{152396}\right] \\ $$

Question Number 152660 Answers: 1 Comments: 0

∫ (x/( (√(1 + x^3 )))) dx

$$\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}\:\:\:+\:\:\:\mathrm{x}^{\mathrm{3}} }}\:\mathrm{dx} \\ $$

Question Number 152653 Answers: 1 Comments: 0

Ω :=∫_0 ^( ∞) (( e^( −x^( 3) ) . sin (x^( 3) ))/x)dx= ((ζ (2 ))/2) m.n...

$$ \\ $$$$\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}^{\:\mathrm{3}} } .\:{sin}\:\left({x}^{\:\mathrm{3}} \:\right)}{{x}}{dx}=\:\frac{\zeta\:\left(\mathrm{2}\:\right)}{\mathrm{2}} \\ $$$$\:{m}.{n}... \\ $$$$ \\ $$

Question Number 152652 Answers: 0 Comments: 1

Question Number 152647 Answers: 1 Comments: 1

Question Number 152631 Answers: 1 Comments: 0

Question Number 152626 Answers: 1 Comments: 0

Solve for real numbers: (1/(x-1)) + (2/(x-2)) + (3/(x-3)) + (4/(x-4)) = 2x^2 -5x-4

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{x}-\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{x}-\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{x}-\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{x}-\mathrm{4}}\:=\:\mathrm{2x}^{\mathrm{2}} -\mathrm{5x}-\mathrm{4} \\ $$

Question Number 152625 Answers: 1 Comments: 2

x^4 +c_3 x^3 +c_2 x^2 +c_1 x+c_0 =0 for c_n ∈R this can have 4 unique zeros ∈R 2 unique zeros + 1 double zero ∈R 2 double zeros ∈R 1 triple + 1 unique zeros ∈R 1 fourfold zero ∈R 2 unique zeros ∈R + 1 pair of complex zeros 1 double zero ∈R + 1 pair of complex zeros 2 pairs of complex zeros 2 double imaginary zeros for given c_n ; can we decide which case we have without solving?

$${x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{for}\:{c}_{{n}} \in\mathbb{R}\:\mathrm{this}\:\mathrm{can}\:\mathrm{have} \\ $$$$\mathrm{4}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:+\:\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{double}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{1}\:\mathrm{triple}\:+\:\mathrm{1}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{1}\:\mathrm{fourfold}\:\mathrm{zero}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R}\:+\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R}\:+\:\mathrm{1}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{zeros} \\ $$$$\mathrm{2}\:\mathrm{double}\:\mathrm{imaginary}\:\mathrm{zeros} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{given}\:{c}_{{n}} ;\:\mathrm{can}\:\mathrm{we}\:\mathrm{decide}\:\mathrm{which}\:\mathrm{case}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{without}\:\mathrm{solving}? \\ $$

Question Number 152617 Answers: 0 Comments: 1