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Yogita Ingle 6 years, 10 months ago
{tex}\begin{array}{l}\frac x8-\;\frac12=\;\frac x{6-2}\\\frac x8-\;\frac12=\;\frac x4\\\frac x8-\frac x4=\;\frac12\\\frac{x\;-\;2x}8\;=\;\frac12\end{array}{/tex}
- x = 4
x = -4
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Yogita Ingle 6 years, 10 months ago
Mean = Total of age of teahers /Total number of student
= 32+ 41 + 28 + 54 + 35 + 26 + 23 + 33+ 38 + 40/10
= 350/10
= 35 years
Posted by Vrinda Bansal 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Values of x and y = 2 and 1 respectively
Solution :
Put x = 2 and y = 1
{2(2)²-5(1)}×{5(2)+2(1)²}
{2*4-5*1}×{5*2+2*1}
{8-5}{10+2}
{3}{12}
{36}
The required value of the polynomial is 36.
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Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}let\;the\;amount\;are\;m\;and\;n\\interest\;I_1=\frac{mx12x1}{100}=\frac{12m}{100}\\I_2=\frac{nx10x1}{100}=\frac{10n}{100}\\I_1+I_2=\frac{12m+10n}{100}=130\\or\;12m+10n=13000\;......(1)\\If\;he\;reverse\;the\;investment\;then\\10m+12n=13400.......(2)\\\\\end{array}{/tex}
{tex}\begin{array}{l}on\;multiplying\;(1)\;by\;5\;and\;(2)\;by\;6\\60m+50n=65000---(3)\\60m+72n=80400----(4)\\subtract\;(3)\;from\;(4)\\22n=15400\\n=\frac{15400}{22}=700\;Rs\\putting\;value\;of\;n\;in\;(1)\\12m+10x700=13000\\12m=13000-7000=6000\\m=500\;Rs\\ans\;is\;Rs\;700\;and\;500\end{array}{/tex}
Posted by Tejashvi 9406205802 6 years, 10 months ago
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Posted by Kirti Gupta 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}2^2+2^2=2\times2+2\times2=4+4\\=8\;\mathrm{ans}\\\end{array}{/tex}
Posted by Vishu Mittal 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
Find coefficient of x
let us find in
Ex (1) : 5x
(1 ) here remove x
(2) then left is 5
(3) This 5 is the coefficient of x
Example 2:
{tex}\begin{array}{l}3\mathrm x^2-12\mathrm{xy}+2\\\mathrm{here}\;\mathrm x\;\mathrm{is}\;\mathrm{in}\;-12\mathrm{xy}\\1\;\mathrm{Remove}\;\mathrm x\\2.\mathrm{we}\;\mathrm{get}\;-12\mathrm y\\3.\;\mathrm{so}\;-12\mathrm y\;\mathrm{is}\;\mathrm{the}\;\mathrm{coeffiiet}\;\mathrm{of}\;\mathrm x\;\\\mathrm{so}\;\mathrm{on}\;..\\\end{array}{/tex}
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Ram Kushwah 6 years, 10 months ago
0Thank You