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ziro4ka [17]
3 years ago
10

In the figure below, if sin x= 5/13, what are cos x and tan x? PLZ HELP FOR THE LOVE OF GOD AND WHTS GOOD IN THE WORLD HELPPP

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

Step-by-step explanation:

cos²<em>x</em> = 1-sin²<em>x</em> = 1-(5/13)² = 144/169

cos<em>x</em> = √(144/169) = 12/13

tan<em>x</em> = sin<em>x</em>/cos<em>x </em>= (5/13)/(12/13) = (5/13)×(13/12) = 5/12

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saveliy_v [14]

5. Answer: see explanation

<u>Step-by-step explanation:</u>

If the roots are m and 3m, then x = m and x = 3m

⇒ x - m = 0    and      x - 3m = 0

⇒ (x - m)(x - 3m) = 0

⇒ x² - 4mx + 3m² = 0

Since x² + px + q = 0   then  p = -4m   and q = 3m²

3p² = 3(-4m)²               16q = 16(3m²)

      = 3(16m²)                      = 48m²

      = 48m²

3p² = 48m² = 16q     ⇒           3p² = 16q

**********************************************************************************************

6. Answer: 8 or 18

<u>Step-by-step explanation:</u>

The Area of the entire rectangle (A = L × w) is 12 × 10 = 120

The Area of the shaded region is 72, so the Area of the non-shaded region is 120 - 72 = 48.

There are two non-shaded triangles.

  • Bigger non-shaded Δ: L = 12-x, w = 10  ⇒ A = \dfrac{10(12-x)}{2}
  • Smaller non-shaded Δ: L = x, w = x ⇒ A=\dfrac{x(x)}{2}

Combine the Areas of both triangles and set it equal to the Area of the non-shaded region:

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Area of ΔBEF:

\text{when x = 4, A =}\ \dfrac{4(4)}{2}=\dfrac{16}{2}=8\\\\\\\text{when x = 6, A =}\ \dfrac{6(6)}{2}=\dfrac{36}{2}=18

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Answer:

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