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Posted by Sonali Singh 6 years, 3 months ago
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The existing Mathematics examination is the Standard Level Examination. Standard-Level and Basic-level Question papers shall be based on the same syllabus. However, the Standard-Level Mathematics assesses higher Mathematical abilities compared to Basic-Level. Accordingly, the difficulty level of Mathematics – ‘Basic’ will be less than that of Mathematics-‘Standard’.
Posted by Kanishka Pal 6 years, 3 months ago
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Posted by Sonal Bajpai 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
The quadratic equation {tex}(k-5)x^2 + 2(k-5)x + 2 = 0{/tex} have equal roots.
⇒ Discriminant (b2 - 4ac) = 0
⇒ {tex}[2(k-5)]^2 - 4(k-5)(2) = 0{/tex}
⇒ {tex}4(k^2 - 10k + 25) - (8k - 40) = 0{/tex}
⇒ {tex}4k^2 - 40k + 100 - 8k + 40 = 0{/tex}
⇒ 4k2 - 48k + 140 = 0
⇒ k2 - 12k + 35 = 0
⇒ (k - 7)(k - 5) = 0
⇒ k = 7 or 5
Posted by Sonal Bajpai 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
{tex}x^2 + x - p(p+1) = 0{/tex}
{tex}x^2 + (p+1)x - px - p(p+1) = 0{/tex}
{tex}x(x+ p + 1) -p(x + p + 1) = 0{/tex}
(x + p + 1)(x - p) = 0
x = - p - 1, p
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Sia ? 6 years, 3 months ago
Let one number be x.
Then, another number = (9 - x) [{tex}\because{/tex} sum of two numbers = 9]
According to the question,
{tex}\begin{array}{l}\frac1x+\;\frac1{9-x}=\;\frac12\\\frac{9\;-x+x}{x(9-x)}\;=\frac12\end{array}{/tex}
⇒ 18 = 9x - x2
{tex}\Rightarrow {x^2} - 9x + 18 = 0 {/tex}
{tex}\Rightarrow {x^2} - 6x - 3x + 18 = 0{/tex}
{tex}\Rightarrow x=3,x=6{/tex}
When, x = 3, then 9 - x = 9 - 3 = 6
When, x = 6, then 9 - x = 9 - 6 = 3
Hence, the numbers are 3 and 6.
Posted by Nitin Singh Nitin Singh 6 years, 3 months ago
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Posted by Mukesh Kumar 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
Let length of rectangle = x units
And width of rectangle = y units
Area of rectangle = length * width = x*y
The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.
So
Decrease the length by 5 unit so new length = x - 5
Increase the width by 3 unit so new width = y + 3
New area is reduced by 9 units
So new area = xy – 9
Plug the value in formula length * width = area we get
(x - 5)(y + 3) = xy - 9
Xy + 3x – 5y – 15 = xy – 9
Subtract xy both side we get
3x - 5y = 6 …(1)
If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units.
Increase the length by 3 unit so new length = x +3
Increase the width by 2 unit so new width = y + 2
New area is increased by 67 units
So new area = xy + 67
Plug the value in formula length * width = area we get
(x +3)(y + 2) = xy + 67
Xy + 2x + 3y + 6 = xy + 67
Subtract xy both side we get
2x + 3y = 61 …(2)*3
3x - 5y = 6 …(1)*2
Cross multiply the coefficient of x we get
6x + 9 y = 183
6x -10y =12
Subtract now we get
19 y = 171
Y = 171/19 = 9
Plug this value of y in equation first we get
2x + 3* 9 = 61
2x = 61 – 27
2x = 34
X = 34/2 = 17
So length is 17 units and width is 9 units
Read more on Brainly.in - https://brainly.in/question/892370#readmore
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Sia ? 6 years, 3 months ago
{tex}3tan\theta = 4{/tex}{tex}\Rightarrow \tan \theta = \frac { 4 } { 3 }{/tex}
Given,
{tex}= \frac { 3 \sin \theta + 2 \cos \theta } { 3 \sin \theta - 2 \cos \theta }{/tex}
{tex}= \frac { 3 \tan \theta + 2 } { 3 \tan \theta - 2 }{/tex}[Dividing numerator and denominator by cos {tex}\theta{/tex}]
{tex}= \frac { \left( 3 \times \frac { 4 } { 3 } + 2 \right) } { \left( 3 \times \frac { 4 } { 3 } - 2 \right) } = \frac { 6 } { 2 } = 3{/tex}
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Yogita Ingle 6 years, 3 months ago
Given √2 is irrational number.
Let √2 = a / b wher a,b are integers b ≠ 0
we also suppose that a / b is written in the simplest form
Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2
∴ 2b2 is divisible by 2
⇒ a2 is divisible by 2
⇒ a is divisible by 2
∴ let a = 2c
a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2
∴ 2c2 is divisible by 2
∴ b2 is divisible by 2
∴ b is divisible by 2
∴a are b are divisible by 2 .
this contradicts our supposition that a/b is written in the simplest form
Hence our supposition is wrong
∴ √2 is irrational number.
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