Question

any maths expert hear I need help I have a pre board sample papers but i don't know answer please help me ​

Answer 1

Step-by-step explanation:

[21]

Let the the point (-1, 6) divides the points (-3, 10) and (6, -8) is the ratio of k : 1

As per SECTION FORMULA

(-1, 6) = $\bigg( \frac{k×6+1×-3}{k+1},\frac{k×-8+1×6}{k+1}\bigg)$

Taking x - coordinates ,

-1 = $\frac{k×6+1×-3}{k+1}$

6k - 3 = -k - 1

7k = 2

k = 2/7

k : 1 = 2 : 7

Hence, the required Ratio is 2 : 7

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[29]

AP → 3, 15, 27, 39

• a = 3
• d = 12
• A_n = a + (n-1)d

A_54 = 3 + (54-1)12

= 3 + 53 × 12

= 639

We need to find the term which is 132 more than A_54 i.e

• A_54 + 132
• 639 + 132
• 771

• A_n = a + (n-1)d
• 771 = 3 + (n-1)12
• 771 - 3 = (n - 1)12
• 12(n - 1) = 768
• n - 1 = 64
• n = 65

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[26]

Quadratic equation has equal roots then b² −4ac=0

In eqn. kx² −2kx+6 = 0, a=k,b=−2k and c=6

Now, (−2k)² −4×k×6=0

• 4k² −24k=0
• 4k(k−6)=0
• k=0 or k=6

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[25]

sinA = p/h = 3/4

• h² = p² + b²
• b² = h² - p²
• b = √(4² - 3²)
• b = √7

cos A = b/h = √7/4

tan A = p/b = 3/√7