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To check whether a point lies inside or outside the circle, we write the equation of circle and substitute the point in the equation.
If the value of the equation is equal to r² then the point lies on the circle.
If the value of equation is less than r² then it lies inside the circle.
And if the value of the equation is greater than r² then it lies outside the circle.
Writing the equation of circle first,
(x -h)² +(x-k)² = r² , where (h,k) is the center of circle and r is the radius of the circle.

Substituting the values we get,

(x -(-2))² +(x-3)² = 5²
Above equation is the equation of the circle.
Now we have to find whether the point lies inside or outside the circle, we substitute the point in the equation of circle.

(2 -(-2))² +(5-3)² = 5²
16+4 = 20 , which is less than r² = 5² =25.

Therefore the point (2,5) lies inside the circle.

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Real number me se ex.1.2

Posted by Vikram Singh Rajput 5 years, 4 months ago

CBSE > Class 10 > Mathematics

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If 3 is the least prime factor of number 'a' and 7 is the least prime factor of number 'b', then what is the least prime factor of (a+b)

Posted by Sneha Sinha 5 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 6 months ago

Then the least prime factor of a is 2; which is not 3 or 7. So a can not be an even number nor can b be an even number. Hence a+b can not be a prime number greater than 2 if the least prime factor of a is 3 and b is 7. Thus the answer is 2.

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Use euclids division lemma to show that the cube of any positive integer is to the form 9m ,9m+1or 9m+8

Posted by Imamuddin Ansari 5 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Akhilesh Achant 5 years, 4 months ago

Let the positive integer (3m)^3,(3m+1)^3,(3m+2)^3 Where m is any positive no. (3m)^3=27 (Where m is any positive no.) 9(3)=9m-(i) (3m+1)^3=27m^3+27m^2+9+1 =9(3m^3+3m^2+1)+1=(9m+1)-(ii) (Where m is any positive integer) (3m+2)^3=27m^3+54m^2+36+8 =9(3m^3+6m^2+4)+8=(9m+8)-(iii) (Where m is any positive integer) From (i),(ii)and(iii)we get, 9m,9m+1,9m+8 Hence prooved (Srividhya)

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https://brainly.in/question/16029041

Posted by Saipreet . 5 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

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If Sn=3n+5n find ap and a20

Posted by Khanak Mandhan 5 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

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If 10th term is 21 and sum of its first 10 terms is 120. Find nth term.

Posted by Khanak Mandhan 5 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Khanak Mandhan 5 years, 4 months ago

Thanks

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Yogita Ingle 5 years, 4 months ago

Let’s consider a to be the first term and d be the common difference.
And we know that, sum of first n terms is:
Sn = n/2(2a + (n − 1)d) and nth term is given by: a_n = a + (n – 1)d
Now, from the question we have S₁₀ = 120
⟹ 120 = 10/2(2a + (10 − 1)d)
⟹ 120 = 5(2a + 9d)
⟹ 24 = 2a + 9d …. (1)
Also given that, a₁₀ = 21
⟹ 21 = a + (10 – 1)d
⟹ 21 = a + 9d …. (2)
Subtracting (2) from (1), we get
24 – 21 = 2a + 9d – a – 9d
⟹a = 3
Now, on putting a = 3 in equation (2), we get
3 + 9d = 21
9d = 18
d = 2
Thus, we have the first term(a) = 3 and the common difference(d) = 2
Therefore, the nth term is given by an = a + (n – 1)d = 3 + (n – 1)2
= 3 + 2n -2
= 2n + 1
Hence, the nth term of the A.P is (an) = 2n + 1.

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15cotA=8,find sinA and secA

Posted by Stunning Prince 5 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Pranathi M 5 years, 4 months ago

15cotA=8 Cot A =8/15 = adjacent side/opposite side Therefore, adjacent side =8 And, opposite side = 15 Hypotenuse = ? By phythagoras theorem, (Hypotenuse )^2 = (side)^2 + (side)^2 (Hypotenuse)^2 = (8)^2 + (15)^2 (Hypotenuse)^2 = 64 +225 (Hypotenuse)^2 = 289 (Hypotenuse) = root 289 Hypotenuse = 17 Therefore, sin A = opp side/ Hypotenuse Sin A = 15/17 Sec A = Hypotenuse/ adjacent side Sec A= 17/8 Hope it helps you ?

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Lucifer ? Morningstar? 5 years, 4 months ago

SinA=15/17 and secA=17/8

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