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Preeti Dabral 3 years, 2 months ago
दिनाँक 13 अगस्त 2023,
प्रिय मित्र दिव्यांशु,
खुश रहो,आज मुझे पता चला कि तुम्हारा चयन हमारे जिले की क्रिकेट टीम में हो गया है और तुम राज्य स्तरीय क्रिकेट टूर्नामेंट के लिए जाने वाले हो। तुम्हारी इस सफलता पर मुझे बड़ी प्रसन्नता हुई।
मैं तुम्हारे सुखद भविष्य की कामना करता हूँ और ईश्वर से प्रार्थना करता हूँ कि तुम्हें यूँ ही निरंतर उन्नति मिलती रहे और तुम सफलता के पथ पर आगे बढ़ते रहो।
आने वाले टूर्नाेमेट तुम सफल होकर आओ, मेरी शुभकामना है।
मेरी इच्छा है कि तुम राज्य स्तरीय टीम में चुने जाओ और उसके बाद राष्ट्रीय टीम में भी चुने जाओ।
मैं तुम्हारे उज्जवल भविष्य की कामना करता हूँ और जिले की टीम में चुने जाने पर पुनः तुम्हें बधाई देता हूँ।
तुम्हारा मित्र,
अंशुल |
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Preeti Dabral 3 years, 2 months ago
Let the other parallel side be x . We know that the area of trapezium is given by the formula :-
Posted by Shivam Singh 3 years, 2 months ago
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Preeti Dabral 3 years, 2 months ago
Cloud computing is on-demand access, via the internet, to computing resources—applications, servers (physical servers and virtual servers), data storage, development tools, networking capabilities, and more—hosted at a remote data center managed by a cloud services provider (or CSP).
Posted by Radhika Rawat 3 years, 2 months ago
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Posted by Soumya Soumya 3 years, 2 months ago
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Preeti Dabral 3 years, 2 months ago
The parallel sides of trapezium are 25 cm and 13 cm and the distance between them is 8 cm.
{tex}\therefore{/tex}Area of trapezium = {tex}\frac{1}{2}{/tex} (sum of parallel sides) {tex}\times{/tex} (perpendicular distance between them)
{tex}=\frac{1}{2}(25+13) \times 8{/tex}
= {tex}38 \times 4{/tex}
= 152 cm2
Posted by Soumya Soumya 3 years, 2 months ago
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Preeti Dabral 3 years, 2 months ago
Given : Length of the rectangular lawn = 25 m
Breadth of the rectangular lawn = 16 m
If lawn is surrounded externally by a path which is 3 m wide. , then length of the path l= 25 m+2(3) m = 31 m
Width of the path b= 16 m +2(3m) = 22 cm
Area of path = l x b - (Area of lawn)
= 31 x 22 - (25 x 16) [area of rectangle = length x breadth]
=682 -400 = 282 sq. m
Rate of levelling = 3.50 per m square
Then, Cost of leveling = (Rate of levelling ) x (Area of path)
= 3.50 x (282)= Rs 987
Hence, the cost of leveling is Rs 987.
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Preeti Dabral 3 years, 2 months ago
Given, {tex}3x -{/tex} {tex}\frac{y+7}{11}{/tex} {tex}+ 2 = 10{/tex}
{tex}\Rightarrow \frac{33 x-(y+7)+22}{11}=10{/tex}
{tex}\Rightarrow{/tex} {tex}33x - y - 7 + 22 = 110{/tex}
{tex}\Rightarrow{/tex} {tex}33x - y = 95{/tex} ..(i)
Also, {tex}2 y+\frac{x+11}{7}{/tex} = 10
{tex}\Rightarrow \quad \frac{14 y+x+11}{7}{/tex} = 10
{tex}\Rightarrow{/tex} {tex}14y + x + 11 = 70{/tex}
{tex}14y + x = 59{/tex}
{tex}14y = 59 - x{/tex}
{tex}\Rightarrow{/tex} y = {tex}\frac{59-x}{14}{/tex}
eq. (i) becomes
33x - {tex}\left(\frac{59-x}{14}\right){/tex} = 95
{tex}\Rightarrow \quad \frac{462 x-59+x}{14}{/tex} = 95
{tex}\Rightarrow{/tex} 463x - 59 = 1330
{tex}\Rightarrow{/tex} x = 3
When x = 3 eq. (i) becomes
33(3) - y = 95
{tex}\Rightarrow{/tex} y =4
Posted by Soumya Soumya 3 years, 2 months ago
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Preeti Dabral 3 years, 2 months ago
Radius of the larger semicircle 14 cm
{tex}\therefore {/tex} Area of the larger semicircle
{tex}= \frac { 1 } { 2 } \pi ( 14 ) ^ { 2 } = \frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 196{/tex}
{tex}= 308 \mathrm { cm } ^ { 2 }{/tex}
Radius of each smaller semicircle = 7 cm
{tex}\therefore {/tex} Area of two smaller semicircles
{tex}= 2 \left[ \frac { 1 } { 2 } \times \pi ( 7 ) ^ { 2 } \right]{/tex}
{tex}= \frac { 22 } { 7 } \times ( 7 ) ^ { 2 } = 154 \mathrm { cm } ^ { 2 }{/tex}
{tex}\therefore {/tex} The area of the shaded region
{tex}= 308 \mathrm { cm } ^ { 2 } + 154 \mathrm { cm } ^ { 2 } = 462 \mathrm { cm } ^ { 2 }{/tex}
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Simran Kumari 3 years, 2 months ago
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