The equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S
To answer the question, we need to know what an equation is
<h3>What is an equation?</h3>
An equation is a mathematical expression that show the relationship between two variables.
Given 9cos(sin¯¹(x)) = √(81 – 81x²), we need to show L.H.S = R.H.S
So, L.H.S = 9cos(sin¯¹(x))
= 9[√{1 - sin²(sin¯¹(x)}] (Since sin²y + cos²y = 1 ⇒ cosy = √[1 - sin²y])
9[√{1 - sin²(sin¯¹(x)}] = √9² × √{1 - sin²(sin¯¹(x)}]
= √[9² × {1 - sin²(sin¯¹(x)}]
= √[81 × {1 - sin²(sin¯¹(x)}]
= √[81 × {1 - x²}] (since sin²(sin¯¹(x) = [sin(sin¯¹(x)]² = x²)
= √(81 – 81x²)
= R.H.S
So, the equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S
Learn more about equations here:
brainly.com/question/2888445
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It will take her 51.4 minutes.
Explanation:
You would do 12/7 and then multiply that answer by 30.
Answer:
(a) By cosine rule, /PQ/2=202+152−2×20×15×cos145°
= 400+225−600cos145°
= 625+600×0.8192
= 491.52+625
/PQ/2=1116.52⟹/PQ/=1116.52−−−−−−√=33.41
≊33km ( to the nearest whole number)
(b) By sine rule,
sin<QPR15=sin14533.41
sin<QPR=15sin14533.41
sin<QPR=15×0.573633.41=0.2575
<QPR=14.92°
∴ Bearing of Q from P = 25° + 14.92° = 39.92°.
≊40° (to nearest whole number)
Step-by-step explanation:
hope it helps
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Function 1 has a maximum at y = 1
Now we need to find the maximum of Function 2 by completing the square:
-x^2 + 2x - 3
= -(x^2 - 2x) - 3
= -(x - 1)^2 +1 - 3
= -(x - 1)^2 - 2
Therefor the turning point is at (1, -2) and the maximum is at y = -2
-2 < 1, therefor Function 1 has the larger maximum
Answer:
y=4x+47
I wanted to wait for the person who originally said the answer but it's been a while and I kinda want the points. Sorry
