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forsale [732]
3 years ago
11

I need help with this question

Mathematics
1 answer:
Novay_Z [31]3 years ago
3 0

Answer:

$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $

$ \frac{-(\sqrt{3} + 1)}{2\sqrt{2}} $

$ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $

Step-by-step explanation:

Given $ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $

(A) $ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4}  + \frac{\pi}{6}) $

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $

$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

(B) Cos(A + B) = CosAcosB - SinASinB

$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $

$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $

$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $

$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}

(C) Tan(A + B) = $ \frac{Sin(A +B)}{Cos(A + B)} $

From the above obtained values this can be calculated and the value is $ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $.

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Explain how you know if an equation has one solution, infinitely many solutions, or no solution
lubasha [3.4K]
No solution would mean that there is no answer to the equation. It is impossible for the equation to be true no matter what value we assign to the variable. Infinite solutions would mean that any value for the variable would make the equation true. :)
8 0
2 years ago
A BASKETBALL PLAYER MAKES A BASKET THROUGH A TEN FOOT HIGH BASKET FROM 20 FEET AWAY. IF THE SHOOTER RELEASES THE BALL FROM 8 FEE
mixas84 [53]

Answer:

The vertical velocity of the ball when released = 10.81 m/s

The vertical velocity of the ball in the basket = 8.81 m/s

Step-by-step explanation:

The given parameters are;

The height of the basket = 10 feet

The horizontal location of the shooter from the basket = 20 feet

The height from which the shooter releases the ball = 8 feet

The horizontal velocity of the ball = 10 ft/s

Therefore, we have;

Time, t = Horizontal distance/(Horizontal velocity)

t = 20/10 = 2 seconds

The vertical velocity is given as follows;

s = u·t - 1/2·g·t²

Where;

u = The vertical velocity

s = The height above the initial height from which the ball is released

s = y - y₀

Where;

y = The final height of the ball in the calculation = The height of the basket = 10 foot

y₀ = The initial height from which the ball is released = 8 feet

∴ s = 10 - 8 = 2 feet

g = The acceleration due to gravity = 9.81 m/s²

Substituting gives;

2 = u × 2 - 1/2 × 9.81× 2²

u = (2 + 1/2 × 9.81× 2²)/2 = 10.81 m/s

The vertical velocity of the ball when released = 10.81 m/s.

The vertical velocity of the ball in the basket is given by the following relation;

v² = u² - 2·g·s

v² = 10.81² - 2×9.81×2 = 77.6161

v = √(77.6161) = 8.81 m/s

The vertical velocity of the ball in the basket = 8.81 m/s

7 0
2 years ago
Helpppppppppppppppppppppppppppppp
n200080 [17]
For this case we first write the equation of which we will use:
 I (db) = 10log (l / l)
 We substitute the value of l.
 We have then:
 l = 10 ^ 8lo
 Substituting in the given equation:
 I (db) = 10log ((10 ^ 8lo) / lo)
 Rewriting:
 I (db) = 10 * log (10 ^ 8)
 I (db) = 80
 Answer:
 I (db) = 80
 option 4
4 0
3 years ago
Help me with this please.
Ira Lisetskai [31]

Answer:

JL = 78

Step-by-step explanation:

MN is a midsegment. Based on the midsegment theorem,

MN = ½(JL)

MN = 5x - 16

JL = 4x + 34

Plug in the value

5x - 16 = ½(4x + 34)

5x - 16 = ½*4x + ½*34

5x - 16 = 2x + 17

Collect like terms

5x - 2x = 16 + 17

3x = 33

Divide both sides by 3

x = 11

✔️JL = 4x + 34

Plug in the value of x

JL = 4(11) + 34

JL = 44 + 34

JL = 78

5 0
3 years ago
Im only have 10 minutes please. Is math​
alukav5142 [94]
Answer: A
Explanation do Pythagorean theorem and find the third side. Then do adjacent/hypotenuse.
7 0
2 years ago
Read 2 more answers
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