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Category: Algebra

To-complete-a-job-24-worker-are-needed-in-35-days-After-they-worked-for-8-days-half-of-the-workers-stopped-working-In-order-for-work-to-be-completed-an-additional-time-of-days-needed-a-

Question Number 133783 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{To}\:\mathrm{complete}\:\mathrm{a}\:\mathrm{job},\:\mathrm{24}\:\mathrm{worker}\:\mathrm{are} \\ $$$$\mathrm{needed}\:\mathrm{in}\:\mathrm{35}\:\mathrm{days}.\:\mathrm{After}\:\mathrm{they}\:\mathrm{worked} \\ $$$$\mathrm{for}\:\mathrm{8}\:\mathrm{days}\:,\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{workers}\: \\ $$$$\mathrm{stopped}\:\mathrm{working}\:.\:\mathrm{In}\:\mathrm{order}\:\mathrm{for} \\ $$$$\mathrm{work}\:\mathrm{to}\:\mathrm{be}\:\mathrm{completed}\:,\:\mathrm{an}\:\mathrm{additional} \\ $$$$\mathrm{time}\:\mathrm{of}\:….\:\mathrm{days}\:\mathrm{needed} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{54}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{38}\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{28}\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{14} \\ $$$$\left(\mathrm{e}\right)\:\mathrm{19}\: \\…

Question-68234

Question Number 68234 by Mikael last updated on 07/Sep/19 Commented by kaivan.ahmadi last updated on 07/Sep/19 $$\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{4}^{{x}} +\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \Rightarrow \\ $$$$\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \left(\mathrm{2}+\mathrm{1}\right)\Rightarrow…

Question-133738

Question Number 133738 by mathlove last updated on 23/Feb/21 Answered by Ar Brandon last updated on 24/Feb/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3x}+\mathrm{1}\right)}{\mathrm{2x}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3x}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}+\epsilon\left(\mathrm{x}\right)}{\mathrm{2x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Commented by…

Prove-by-contradiction-that-there-are-no-whole-number-solutions-x-y-z-to-the-equation-z-2-x-2-y-2-where-both-x-and-y-are-odd-

Question Number 2655 by Yozzi last updated on 24/Nov/15 $${Prove}\:{by}\:{contradiction}\:{that}\:{there} \\ $$$${are}\:{no}\:{whole}\:{number}\:{solutions}\:\left({x},{y},{z}\right) \\ $$$${to}\:{the}\:{equation}\:{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${where}\:{both}\:{x}\:{and}\:{y}\:{are}\:{odd}. \\ $$ Answered by prakash jain last…

n-lines-are-drawn-inside-a-circle-in-such-a-way-that-the-circle-has-been-divided-in-maximum-number-of-parts-Determine-this-maximum-number-

Question Number 2642 by Rasheed Soomro last updated on 24/Nov/15 $${n}\:{lines}\:{are}\:{drawn}\:{inside}\:{a}\:{circle}\:{in}\:{such}\:{a}\:{way}\:{that}\: \\ $$$${the}\:{circle}\:{has}\:{been}\:{divided}\:{in}\:{maximum}\:{number}\:{of} \\ $$$${parts}.\:{Determine}\:{this}\:{maximum}\:{number}. \\ $$ Commented by RasheedAhmad last updated on 24/Nov/15 $$\bullet{One}\:{line}\:{can}\:{divide}\:{the}\:{circle}…

In-my-textbook-its-written-In-applying-the-nth-term-test-we-can-see-that-n-1-1-n-1-diverges-because-lim-n-1-n-1-does-not-exist-But-then-why-n-1-1-n-1-1-n-2-

Question Number 68161 by Learner-123 last updated on 06/Sep/19 $${In}\:{my}\:{textbook}\:{its}\:{written}: \\ $$$${In}\:{applying}\:{the}\:{nth}−{term}\:{test}\:{we}\: \\ $$$${can}\:{see}\:{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{diverges}\:{because}\: \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{does}\:{not}\:{exist}. \\ $$$${But}\:{then}\:{why}\:\underset{{n}=\mathrm{1}}…

The-sums-of-the-first-n-terms-of-two-AP-s-are-in-the-ratio-3n-31-5n-3-Show-that-their-9-th-terms-are-equal-

Question Number 2619 by Rasheed Soomro last updated on 23/Nov/15 $${The}\:{sums}\:{of}\:{the}\:{first}\:\:{n}\:\:\:{terms}\:{of}\:{two}\:{AP}\:'{s}\:{are} \\ $$$${in}\:{the}\:{ratio}\:\:\mathrm{3}{n}+\mathrm{31}\::\:\:\mathrm{5}{n}−\mathrm{3}\:.\:{Show}\:{that}\:{their}\:\mathrm{9}^{{th}} \:{terms} \\ $$$${are}\:{equal}. \\ $$ Commented by Yozzi last updated on 24/Nov/15…

Question-133664

Question Number 133664 by shaker last updated on 23/Feb/21 Answered by liberty last updated on 23/Feb/21 $$\mathrm{partial}\:\mathrm{fraction} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{\mathrm{A}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{B}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Leftrightarrow\:\mathrm{1}=\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}}…