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AlgebraQuestion and Answers: Page 251

Question Number 104369 Answers: 1 Comments: 0

Question Number 104326 Answers: 1 Comments: 2

Suppose you had x rupees. Your father gave you more 6 rupees. You gave y rupees to your brother. Now how many rupees you have? You will buy three shirts with the rupees that you have now. Write the cost of each shirt.

Question Number 104297 Answers: 1 Comments: 0

((3a^2 b^3 c)/(5b^4 c))+((6xy^3 z^(16) )/(10x^2 y^2 z^(10) )) = ? Can you solve this?

$$\frac{\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}}{\mathrm{5}{b}^{\mathrm{4}} {c}}+\frac{\mathrm{6}{xy}^{\mathrm{3}} {z}^{\mathrm{16}} }{\mathrm{10}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{10}} }\:=\:? \\ $$$$\boldsymbol{\mathrm{Can}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{this}}? \\ $$

Question Number 104294 Answers: 1 Comments: 0

1. ((1(1/2)+2(6/7))/(2(2/3)−3(4/5)))=? 2. 4×Π×Π=? 3. Transfer into fractions: 5.8^. 9^. , 9.6^. , 78.57^. 8^.

$$\mathrm{1}.\:\:\frac{\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\frac{\mathrm{6}}{\mathrm{7}}}{\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{3}\frac{\mathrm{4}}{\mathrm{5}}}=? \\ $$$$\mathrm{2}.\:\:\mathrm{4}×\Pi×\Pi=? \\ $$$$\mathrm{3}.\:\:\boldsymbol{\mathrm{Transfer}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{fractions}}: \\ $$$$\:\:\:\:\:\:\mathrm{5}.\overset{.} {\mathrm{8}}\overset{.} {\mathrm{9}},\:\mathrm{9}.\overset{.} {\mathrm{6}},\:\mathrm{78}.\mathrm{5}\overset{.} {\mathrm{7}}\overset{.} {\mathrm{8}} \\ $$

Question Number 104278 Answers: 1 Comments: 0

x−(−(−x+(−x+x)))= ?

$${x}−\left(−\left(−{x}+\left(−{x}+{x}\right)\right)\right)=\:? \\ $$

Question Number 104260 Answers: 0 Comments: 1

You bought 3g rice, 5g flour in a market. At first you had 500 rupees. Now you have 300 rupees. How much rupees you wasted? Suppose you distribute the 300 rupees among your 4 sons. Now how much rupees does your one son get?

$$\mathrm{You}\:\mathrm{bought}\:\mathrm{3g}\:\mathrm{rice},\:\mathrm{5g}\:\mathrm{flour}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{market}.\:\mathrm{At}\:\mathrm{first}\:\mathrm{you}\:\mathrm{had}\:\mathrm{500}\:\mathrm{rupees}. \\ $$$$\mathrm{Now}\:\mathrm{you}\:\mathrm{have}\:\mathrm{300}\:\mathrm{rupees}.\:\mathrm{How}\:\mathrm{much} \\ $$$$\mathrm{rupees}\:\mathrm{you}\:\mathrm{wasted}?\:\mathrm{Suppose}\:\mathrm{you} \\ $$$$\mathrm{distribute}\:\mathrm{the}\:\mathrm{300}\:\mathrm{rupees}\:\mathrm{among} \\ $$$$\mathrm{your}\:\mathrm{4}\:\mathrm{sons}.\:\mathrm{Now}\:\mathrm{how}\:\mathrm{much}\:\mathrm{rupees} \\ $$$$\mathrm{does}\:\mathrm{your}\:\mathrm{one}\:\mathrm{son}\:\mathrm{get}? \\ $$

Question Number 104253 Answers: 1 Comments: 1

When a∗b= ((a+b)/(a−b)) then what is the answer of 2∗3×9∗10 ?

$$\mathrm{When}\:{a}\ast{b}=\:\frac{{a}+{b}}{{a}−{b}}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{answer}\:\mathrm{of}\:\mathrm{2}\ast\mathrm{3}×\mathrm{9}\ast\mathrm{10}\:? \\ $$

Question Number 104251 Answers: 1 Comments: 0

If x=8 and y=5 then what is the answer of ((x+y+6)/(x^2 −y^3 )) ?

$$\mathrm{If}\:{x}=\mathrm{8}\:\mathrm{and}\:{y}=\mathrm{5}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{answer}\:\mathrm{of}\:\:\frac{{x}+{y}+\mathrm{6}}{{x}^{\mathrm{2}} −{y}^{\mathrm{3}} }\:? \\ $$

Question Number 104246 Answers: 3 Comments: 0

(1−(1/3))(1−(1/4))(1−(1/5))....(1−(1/(99))) =?

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)....\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{99}}\right)\:=? \\ $$

Question Number 104243 Answers: 2 Comments: 0

prove that π=2×(2/(√2))×(2/(√(2+(√2))))×(2/(√(2+(√(2+(√2))))))×.....

$${prove}\:{that} \\ $$$$\pi=\mathrm{2}×\frac{\mathrm{2}}{\sqrt{\mathrm{2}}}×\frac{\mathrm{2}}{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}×\frac{\mathrm{2}}{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}×..... \\ $$

Question Number 104270 Answers: 1 Comments: 0

sgn(∣x∣)=?

$${sgn}\left(\mid{x}\mid\right)=? \\ $$

Question Number 104235 Answers: 1 Comments: 0

If you are given a triangle with side length 15 , 20 and 25. what is the triangle′s shortest altitude?

$${If}\:{you}\:{are}\:{given}\:{a}\:{triangle} \\ $$$${with}\:{side}\:{length}\:\mathrm{15}\:,\:\mathrm{20}\:{and} \\ $$$$\mathrm{25}.\:{what}\:{is}\:{the}\:{triangle}'{s} \\ $$$${shortest}\:{altitude}? \\ $$

Question Number 104201 Answers: 0 Comments: 0

If a^2 + b^2 + c^2 = 1 and b + ic = (1 + a)z, then show that ((a + ib)/(i + c)) = ((1 + iz)/(1 − iz))

$$\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\mathrm{1}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{b}\:\:+\:\:\mathrm{ic}\:\:=\:\:\left(\mathrm{1}\:\:+\:\:\mathrm{a}\right)\mathrm{z}, \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\:\:\:\:\:\frac{\mathrm{a}\:\:+\:\:\mathrm{ib}}{\mathrm{i}\:\:+\:\:\mathrm{c}}\:\:\:=\:\:\:\frac{\mathrm{1}\:\:+\:\:\mathrm{iz}}{\mathrm{1}\:\:−\:\:\mathrm{iz}} \\ $$

Question Number 104190 Answers: 3 Comments: 1

integers x,y satisfy 2x+15y=2019. find the minimum of ∣y−x∣.

$${integers}\:{x},{y}\:{satisfy}\:\mathrm{2}{x}+\mathrm{15}{y}=\mathrm{2019}. \\ $$$${find}\:{the}\:{minimum}\:{of}\:\mid{y}−{x}\mid. \\ $$

Question Number 104115 Answers: 1 Comments: 0

Question Number 104086 Answers: 1 Comments: 3

Solve: log_r 8 + log_3 p = 5 ..... (i) r + p = 11 ..... (ii)

$$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{log}_{\mathrm{r}} \mathrm{8}\:\:\:+\:\:\:\mathrm{log}_{\mathrm{3}} \mathrm{p}\:\:\:=\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:.....\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{r}\:\:\:+\:\:\mathrm{p}\:\:\:=\:\:\mathrm{11}\:\:\:\:\:\:\:\:\:.....\:\left(\mathrm{ii}\right) \\ $$

Question Number 104062 Answers: 3 Comments: 4

{ (((x/y) + (y/x) = ((13)/6))),((x+y = 5)) :} find the solution

$$\begin{cases}{\frac{{x}}{{y}}\:+\:\frac{{y}}{{x}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}}\\{{x}+{y}\:=\:\mathrm{5}}\end{cases} \\ $$$${find}\:{the}\:{solution} \\ $$

Question Number 104037 Answers: 4 Comments: 0

(1) { ((x^3 +y^6 = 91)),((x+y^2 = 7 )) :} find x−y^6 . (2) 2a+(2/a) = 8 ⇒ ((a^6 +1)/a^3 ) ?

$$\left(\mathrm{1}\right)\begin{cases}{{x}^{\mathrm{3}} +{y}^{\mathrm{6}} \:=\:\mathrm{91}}\\{{x}+{y}^{\mathrm{2}} \:=\:\mathrm{7}\:}\end{cases} \\ $$$${find}\:{x}−{y}^{\mathrm{6}} \:. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}{a}+\frac{\mathrm{2}}{{a}}\:=\:\mathrm{8}\:\Rightarrow\:\frac{{a}^{\mathrm{6}} +\mathrm{1}}{{a}^{\mathrm{3}} }\:? \\ $$

Question Number 104023 Answers: 0 Comments: 1

Question Number 103983 Answers: 0 Comments: 0

If α and β are two unequal angle, which satisfy the equation, a cos(α) + b sin(β) = c, show that (i) sin(((α + β)/2)) sec(((α − β)/2)) = (b/c) (ii) tan((α/2)) tan((β/2)) = ((c − a)/(c + a))

$$\mathrm{If}\:\:\alpha\:\:\mathrm{and}\:\:\beta\:\:\mathrm{are}\:\mathrm{two}\:\mathrm{unequal}\:\mathrm{angle},\:\mathrm{which}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}, \\ $$$$\:\:\:\:\mathrm{a}\:\mathrm{cos}\left(\alpha\right)\:\:+\:\:\mathrm{b}\:\mathrm{sin}\left(\beta\right)\:\:=\:\:\mathrm{c},\:\:\:\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\:\:\:\mathrm{sin}\left(\frac{\alpha\:\:+\:\beta}{\mathrm{2}}\right)\:\mathrm{sec}\left(\frac{\alpha\:\:−\:\:\beta}{\mathrm{2}}\right)\:\:=\:\:\frac{\mathrm{b}}{\mathrm{c}} \\ $$$$\left(\mathrm{ii}\right)\:\:\:\:\:\mathrm{tan}\left(\frac{\alpha}{\mathrm{2}}\right)\:\mathrm{tan}\left(\frac{\beta}{\mathrm{2}}\right)\:\:=\:\:\frac{\mathrm{c}\:\:−\:\:\mathrm{a}}{\mathrm{c}\:\:+\:\:\mathrm{a}} \\ $$

Question Number 103921 Answers: 1 Comments: 0

Prove that ∀ x ∈ R^ , ∣ cos x ∣ ≤ 1 − sin^2 x

$$\mathrm{Prove}\:\mathrm{that}\:\forall\:{x}\:\in\:\bar {\mathbb{R}}\:,\:\mid\:\mathrm{cos}\:{x}\:\mid\:\leqslant\:\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$

Question Number 103914 Answers: 2 Comments: 2

(2+(√3))^x^2 + (2−(√3))^x^2 = 4

$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:=\:\mathrm{4}\: \\ $$

Question Number 103888 Answers: 2 Comments: 0

find all such numbers: if we make its last digit, say k, as its first digit, the number becomes k times large as before. (□□...□k)→(k□□...□)=k×(□□...□k)

$${find}\:{all}\:{such}\:{numbers}: \\ $$$${if}\:{we}\:{make}\:{its}\:{last}\:{digit},\:{say}\:{k},\:{as}\:{its} \\ $$$${first}\:{digit},\:{the}\:{number}\:{becomes}\:{k} \\ $$$${times}\:{large}\:{as}\:{before}. \\ $$$$\left(\Box\Box...\Box{k}\right)\rightarrow\left({k}\Box\Box...\Box\right)={k}×\left(\Box\Box...\Box{k}\right) \\ $$

Question Number 103804 Answers: 0 Comments: 0

Solve for r ((h−p^2 )/(p−r))=(1/(2p)) , ((h−q)/(q^2 −r))= 2q (h−p^2 )^2 +(p−r)^2 =r^2 (h−q)^2 +(q^2 −r)^2 =r^2

$$\:\:\:\boldsymbol{{S}}{olve}\:{for}\:\boldsymbol{{r}} \\ $$$$\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{p}}−\boldsymbol{{r}}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{p}}}\:\:\:\:,\:\:\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{q}}}{\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}}=\:\mathrm{2}\boldsymbol{{q}} \\ $$$$\left(\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\:\left(\boldsymbol{{h}}−\boldsymbol{{q}}\right)^{\mathrm{2}} +\left(\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$

Question Number 103736 Answers: 0 Comments: 0

Solve: a^2 + c^2 = 196 ... (i) b^2 + (c − a)^2 = 169 ... (ii) c^2 + (b − c)^2 = 225 ... (iii)

$$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\:\mathrm{196}\:\:\:\:\:\:...\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{c}\:\:−\:\:\mathrm{a}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{169}\:\:\:\:\:...\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{b}\:\:−\:\:\mathrm{c}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{225}\:\:\:\:\:...\:\left(\mathrm{iii}\right) \\ $$

Question Number 103685 Answers: 1 Comments: 0

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