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Yogita Ingle 5 years, 3 months ago
An equation that can be put in the form ax + by + c = 0, where a, b and c are real numbers and a, b not equal to zero is called a linear equation in two variables namely x and y. The solution for such an equation is a pair of values, one for x and one for y which further makes the two sides of an equation equal.
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Yogita Ingle 5 years, 3 months ago
Let the base of the triangle 
then altitude of the triangle 
Hypotenuse = 13 cm
By Pythagoras theorem
(Hypotenuse) 2 = (Base)2 + (altitude)2
Length can not be negative hence we choose x = 12 cm
Altitude = x - 7 = 12 - 7 = 5 cm
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Sakshi Singh 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
The number of roots of a polynomial equation is equal to its degree. Hence, a quadratic equation has 2 roots. Let α and β be the roots of the general form of the quadratic equation :ax2 + bx + c = 0. We can write:
α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a
Here a, b, and c are real and rational. Hence, the nature of the roots α and β of equation ax2 + bx + c = 0 depends on the quantity or expression (b2 – 4ac) under the square root sign. We say this because the root of a negative number can’t be any real number. Say x2 = -1 is a quadratic equation. There is no real number whose square is negative. Therefore for this equation, there are no real number solutions.
Hence, the expression (b2 – 4ac) is called the discriminant of the quadratic equation ax2 + bx + c = 0. Its value determines the nature of roots as we shall see. Depending on the values of the discriminant, we shall see some cases about the nature of roots of different quadratic equations
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Prakash Singh 5 years, 3 months ago
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Posted by Saurav Kumar 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
x2-5x+k
Here, a=1, b=-5 and c=k
Now, α+ β = -b/a= -(-5)/1= 5
α*β = c/a= k/7= k
Now,α - β =1
Squaring both sides, we get,
(α - β)2=12
⇒ α2 + β2 - 2αβ = 1
⇒ (α2 + β2 + 2αβ) - 4αβ = 1
⇒ (α +β)2 -4αβ =1
⇒ (5)2-4k=1
⇒ -4k= 7-25
⇒ -4k= -24
⇒ k=6 So the value of k is 6.
Posted by Manuj Arela 5 years, 3 months ago
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Posted by Ganga Enterprises 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
Total pens=1001
Total pencils=910
we need to find maximum no.of students among whom 1001 pens and 910 pencils can be distributed in such a way that each students get same no.of pens and pencils.
Then we need to find HCF of 1001 and 910
Prime factorization of,
1001=7×11×13
910=2×5×7×13
HCF=product of commom prime factor of least power
HCF=7×13=91
Here HCF of 1001 and 910 is 91.
Hence among 91 students 1001 pens and 910 pencils can be distributed such that each student get same no.of pens and pencils.
Posted by Ganga Enterprises 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively.
Hence, using LCM of the given numbers which is 120 we conclude that bell will toll together after 120 seconds = 2 minutes.
In 30 minutes they will toll together in 30/2 = 15 and 1 (at the starting).
Hence, total number of bells together is 15 + 1 = 16
Posted by Nivi :) 5 years, 3 months ago
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Priyal Gupta 5 years, 3 months ago
Posted by Nivi :) 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
Using Euclid's division lemma
hence HCF is 2. Now starts from second last ewuation
so m=4
n= -59
Posted by Nivi :) 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
LCM × HCF = Product of the Numbers
LCM × 9 = 306 × 630
LCM = (306 × 630)/9
LCM = 306 × 70
LCM= 21420
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