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AlgebraQuestion and Answers: Page 91

Question Number 179100    Answers: 4   Comments: 2

if x^2 +y^2 +xy=5, find the range of x^2 +y^2 −xy. (x,y ∈R)

$${if}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{5},\:{find}\:{the}\:{range}\:{of} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}. \\ $$$$\left({x},{y}\:\in\mathbb{R}\right) \\ $$

Question Number 179095    Answers: 2   Comments: 1

Question Number 179090    Answers: 1   Comments: 1

Given a>0, b>0, c>2, a+b=1. find the minimum value of ((3ac)/b)+(c/(ab))+(6/(c−2)).

$$\mathrm{Given}\:{a}>\mathrm{0},\:{b}>\mathrm{0},\:{c}>\mathrm{2},\:{a}+{b}=\mathrm{1}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\:\frac{\mathrm{3}{ac}}{{b}}+\frac{{c}}{{ab}}+\frac{\mathrm{6}}{{c}−\mathrm{2}}. \\ $$

Question Number 179064    Answers: 1   Comments: 3

Question Number 179063    Answers: 1   Comments: 2

why is not a polynomial (√(25x^8 )) ?

$${why}\:{is}\:{not}\:{a}\:{polynomial}\:\sqrt{\mathrm{25}{x}^{\mathrm{8}} }\:\:? \\ $$

Question Number 179062    Answers: 1   Comments: 1

why is not it a polynomial ∣10−2y∣?

$${why}\:{is}\:{not}\:{it}\:{a}\:{polynomial}\:\mid\mathrm{10}−\mathrm{2}{y}\mid? \\ $$

Question Number 179058    Answers: 1   Comments: 0

f(x)=(((x−2)/(x+1)))^(1/3) Dom_(f(x)) =?

$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}\:\:\:\:\:\:\:{Dom}_{{f}\left({x}\right)} =? \\ $$

Question Number 179031    Answers: 0   Comments: 1

prove that sgn(0)=0

$${prove}\:{that}\:{sgn}\left(\mathrm{0}\right)=\mathrm{0} \\ $$

Question Number 179025    Answers: 0   Comments: 3

2^(10x) −x^5 −4=0

$$\mathrm{2}^{\mathrm{10}{x}} −{x}^{\mathrm{5}} −\mathrm{4}=\mathrm{0} \\ $$

Question Number 179023    Answers: 0   Comments: 0

Draw an electrical network (a) p∧(q∨r) (b)(∼p∧∼q)∨(∼p∧q)∨(p∧∼q) (c) p↔q

$$\mathrm{Draw}\:\mathrm{an}\:\mathrm{electrical}\:\mathrm{network} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{p}\wedge\left(\mathrm{q}\vee\mathrm{r}\right) \\ $$$$\left(\mathrm{b}\right)\left(\sim\mathrm{p}\wedge\sim\mathrm{q}\right)\vee\left(\sim\mathrm{p}\wedge\mathrm{q}\right)\vee\left(\mathrm{p}\wedge\sim\mathrm{q}\right) \\ $$$$\left(\mathrm{c}\right)\:\mathrm{p}\leftrightarrow\mathrm{q} \\ $$

Question Number 179018    Answers: 3   Comments: 0

Question Number 178996    Answers: 2   Comments: 0

Question Number 179001    Answers: 0   Comments: 1

In △ABC ∠BAC = 90° and AB = ((BC)/2). ∠ACB = ?

$$\mathrm{In}\:\bigtriangleup\mathrm{ABC}\:\angle\mathrm{BAC}\:=\:\mathrm{90}°\:\mathrm{and}\:\mathrm{AB}\:=\:\frac{\mathrm{BC}}{\mathrm{2}}. \\ $$$$\angle\mathrm{ACB}\:=\:? \\ $$

Question Number 178957    Answers: 1   Comments: 0

Find the recurring factors of the polynomial: 1. f(x)=x^5 −6x^4 +16x^3 −24x^2 +20x−8 2. f(x)=x^4 −x^3 −30x^2 −7x−56

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{recurring}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}: \\ $$$$\mathrm{1}.\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{5}} −\mathrm{6x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{24x}^{\mathrm{2}} +\mathrm{20x}−\mathrm{8} \\ $$$$\mathrm{2}.\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} −\mathrm{30x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{56} \\ $$

Question Number 178923    Answers: 2   Comments: 3

Question Number 178915    Answers: 0   Comments: 0

Question Number 178912    Answers: 1   Comments: 0

1×2+2×3+3×4+...+99×100=?

$$\mathrm{1}×\mathrm{2}+\mathrm{2}×\mathrm{3}+\mathrm{3}×\mathrm{4}+...+\mathrm{99}×\mathrm{100}=? \\ $$

Question Number 178867    Answers: 2   Comments: 0

Question Number 178818    Answers: 1   Comments: 0

If a,b,c,d∈[1,2] then prove that: (a + b + c + d + e)((1/a) + (1/b) + (1/c) + (1/d) + (1/e)) ≤ 28 When equality holds?

$$\mathrm{If}\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in\left[\mathrm{1},\mathrm{2}\right]\:\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d}\:+\:\mathrm{e}\right)\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:+\:\frac{\mathrm{1}}{\mathrm{d}}\:+\:\frac{\mathrm{1}}{\mathrm{e}}\right)\:\leqslant\:\mathrm{28} \\ $$$$\mathrm{When}\:\mathrm{equality}\:\mathrm{holds}? \\ $$

Question Number 178810    Answers: 1   Comments: 1

Question Number 178794    Answers: 0   Comments: 0

i i

$${i} \\ $$$${i} \\ $$

Question Number 178793    Answers: 2   Comments: 0

find Σ_(n=0) ^∞ (1/((3n)!)) = 1+(1/(3!))+(1/(6!))+(1/(9!))+...

$$\mathrm{find}\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{3}{n}\right)!}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{6}!}+\frac{\mathrm{1}}{\mathrm{9}!}+... \\ $$

Question Number 178777    Answers: 1   Comments: 6

A=(1/(2018))+(1/(2019))+...+(1/(2050)) find the integer part of (1/A).

$${A}=\frac{\mathrm{1}}{\mathrm{2018}}+\frac{\mathrm{1}}{\mathrm{2019}}+...+\frac{\mathrm{1}}{\mathrm{2050}} \\ $$$${find}\:{the}\:{integer}\:{part}\:{of}\:\frac{\mathrm{1}}{{A}}. \\ $$

Question Number 178747    Answers: 1   Comments: 0

{ ((3x=2 (mod 5))),((3x=4 (mod 7) )),((3x=6 (mod 11))) :} x=?

$$\:\:\begin{cases}{\mathrm{3x}=\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{3x}=\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:}\\{\mathrm{3x}=\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\end{cases} \\ $$$$\:\mathrm{x}=? \\ $$

Question Number 178715    Answers: 0   Comments: 0

The solution of 2≤ax^2 +bx+c≤3 is [2, 3] 1) if a>0, ax^2 +(b−3)x−c≤0 has and only has 10 integer solutions. find the range of a. 2) find x: ax^2 +(b−1)x+5<0

$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{2}\leqslant{ax}^{\mathrm{2}} +{bx}+{c}\leqslant\mathrm{3}\:\mathrm{is}\:\left[\mathrm{2},\:\mathrm{3}\right] \\ $$$$\left.\mathrm{1}\right)\:\mathrm{if}\:{a}>\mathrm{0},\:{ax}^{\mathrm{2}} +\left({b}−\mathrm{3}\right){x}−{c}\leqslant\mathrm{0}\:\mathrm{has}\:\mathrm{and}\:\mathrm{only}\:\mathrm{has}\:\mathrm{10}\:\mathrm{integer}\:\mathrm{solutions}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:{x}:\:{ax}^{\mathrm{2}} +\left({b}−\mathrm{1}\right){x}+\mathrm{5}<\mathrm{0} \\ $$

Question Number 178697    Answers: 1   Comments: 3

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