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Ratling [72]
2 years ago
15

If a = 5+2√6 then find a²-1÷a²

Mathematics
1 answer:
Umnica [9.8K]2 years ago
5 0

Answer:

a²+1/a² = 7

Step-by-step explanation:

Given, a = (3+√5)/2

We have to find the value of a² + 1/a².

1/a = 1/((3+√5)/2) = 2/(3+√5)

By taking conjugate,

2/(3+√5) = 2/(3+√5) × (3-√5)/(3-√5)

= 2(3-√5) / (3+√5)(3-√5)

By using algebraic identity,

(a - b)(a + b) = a² - b²

(3+√5)(3-√5) = (3)² - (√5)²

= 9 - 5

= 4

2(3-√5) / (3+√5)(3-√5) = 2(3-√5) / 4

So, 1/a = (3-√5)/2

Now, a² = [(3+√5)/2]²

= (3+√5)²/4

By using algebraic identity,

(a + b)² = a² + 2ab + b²

(3+√5)² = (3)² + 2(3)(√5) + (√5)²

= 9 + 6√5 + 5

= 14 + 6√5

a² = (14+6√5)/4

= 2(7+3√5)/4

a² = (7+3√5)/2

Now, 1/a² = [(3-√5)/2]²

= (3-√5)²/4

By using algebraic identity,

(a - b)² = a² - 2ab + b²

(3-√5)² = (3)² - 2(3)(√5) + (√5)²

= 9 - 6√5 + 5

= 14 - 6√5

1/a² = (14-6√5)/4

1/a² = (7-3√5)/2

a²+1/a² = (7+3√5)/2 + (7-3√5)/2

= 7/2 + 3√5/2 + 7/2 - 3√5/2

= 7/2 + 7/2

= 7

Therefore, a²+1/a² = 7

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The two men measure the angle of elevation to the highest point on the rock to be 22.6°. They
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Answer:

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Step-by-step explanation:

<em>Refer to attached picture</em>

  • The height  = h
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<u>Use tangent to solve</u>

  • tan 22.6° = h/(d + 17) ⇒ h = (d + 17) tan 22.6° ⇒ h = 0.42(d + 17)
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<u>Compare the equations above and find the value of d:</u>

  • 0.42(d + 17) = 0.79d
  • 0.42d + 7.14 = 0.79d
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3 years ago
If a cannonball is shot directly upward with a velocity of 160 ft per​ second, its height above the ground after t seconds is gi
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Step-by-step explanation:

If a cannonball is shot directly upward with a velocity of 160 ft per​ second. Its height as a function of time is given by :

h(t)=160t-16t^2 .......(1)

t is in seconds

(a) Velocity is given by :

v=\dfrac{dh}{dt}\\\\v=\dfrac{d(160t-16t^2)}{dt}\\\\v=(160-32t)\ ft/s  

Acceleration is given by :

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(160-32t)}{dt}\\\\a=-32\ ft/s^2

(b) For maximum height put \dfrac{dh}{dt}=0

i.e.

160-32t=0\\\\t=5\ s

Put t = 5 s in equation (1). So,

h(5)=160t-16t^2\\\\h(5)=160(5)-16(5)^2\\\\h(5)=400\ ft

(c) When the ball reaches ground, its height is equal to 0. So,

h(t) = 0

160t-16t^2=0\\\\t(160-16t)=0\\\\t=0,t=10\ s

Hence, this is the required solution.

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