<?xml version="1.0" encoding="UTF-8"?>
<!-- generator="FeedCreator 1.8" -->
<?xml-stylesheet href="https://www.mathcity.org/lib/exe/css.php?s=feed" type="text/css"?>
<rss version="2.0">
    <channel xmlns:g="http://base.google.com/ns/1.0">
        <title>MathCity.org</title>
        <description>Merging man &amp; maths</description>
        <link>https://www.mathcity.org/</link>
        <lastBuildDate>Fri, 05 Jun 2026 17:12:43 +0000</lastBuildDate>
        <generator>FeedCreator 1.8</generator>
        <image>
            <url>https://www.mathcity.org/_media/logo.svg</url>
            <title>MathCity.org</title>
            <link>https://www.mathcity.org/</link>
        </image>
        <item>
            <title>Question 7 &amp; 8 Review Exercise 6</title>
            <link>https://www.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p5</link>
            <description>Question 7 &amp; 8 Review Exercise 6

Solutions of Question 7 &amp; 8 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$P(A)=0.8, P(B)=0.5$$P(B / A)=0.4$$P(A \cap B)$\begin{align}
P(B \mid A)&amp;=\dfrac{P(A \cap B)}{P(A)} \\
\Rightarrow P(A \cap B)&amp;=P(B \mid A) \cdot P(A)\\
&amp;=0.4 \times 0.8=0.32\end{align}$P(A)=0.8, P(B)=0.5$$P(B / A)=0.4$$P(A …</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Thu, 01 Feb 2024 02:36:01 +0000</pubDate>
        </item>
        <item>
            <title>Question 9 &amp; 10 Review Exercise 6</title>
            <link>https://www.mathcity.org/math-11-kpk/sol/unit06/re-ex6-p6</link>
            <description>Question 9 &amp; 10 Review Exercise 6

Solutions of Question 9 &amp; 10 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$2,3,0,3,4,2,3$$1$$=100,0000$$$=\dfrac{7 !}{3 ! \cdot 2 !}=420 $$$1$$0$$7$$0$$$=\dfrac{6 !}{2 ! 3 !}=60 $$$1$$420-50=360$$n$$n$$(n-1)$$(n - 1)$$(n-1)$$(n-2) !$$2$$2 !$$n$$$(n-2) ! \cdot 2 !=2(n-2) ! $$</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Thu, 01 Feb 2024 02:36:03 +0000</pubDate>
        </item>
        <item>
            <title>Question 15 &amp; 16 Exercise 4.5</title>
            <link>https://www.mathcity.org/math-11-kpk/sol/unit04/ex4-5-p10</link>
            <description>Question 15 &amp; 16 Exercise 4.5

Solutions of Question 15 &amp; 16 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$2$$4$$15^{\text {th }}$$a_1=R s .1$$a_2=R s .2$$a_3=R s .4$$1,2,4,8, \ldots$$a_1=1 . \quad r=2 . \quad n=15$$a_n=a_1 r^{n-1}$$15^{1 / 2}$$$a_{15}=a_1 r^{14} $$$$a_{15}=1 .(2)^{1 4}=R s .16384 $$$$S_{30}=\dfrac{a_1(r^{30}-1)}{r-1} $$$r-2$$a_1=1$\begi…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Fri, 05 Jan 2024 17:30:06 +0000</pubDate>
        </item>
        <item>
            <title>Question 6, Exercise 10.1</title>
            <link>https://www.mathcity.org/math-11-kpk/sol/unit10/ex10-1-p6</link>
            <description>Question 6, Exercise 10.1

Solutions of Question 1 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\cos \alpha =2{{\cos }^{2}}\dfrac{\alpha }{2}-1=1-2{{\sin }^{2}}\dfrac{\alpha }{2}$\begin{align}\cos \alpha &amp;=\cos 2\dfrac{\alpha }{2}\\
&amp;={{\cos }^{2}}\dfrac{\alpha }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}\\ 
&amp;={{\cos }^{2}}…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Tue, 05 Dec 2023 02:45:42 +0000</pubDate>
        </item>
    </channel>
</rss>
