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Natalija [7]
3 years ago
6

Order the following sets of numbers from least to greatest 2.1 notation bar,-2.1,2 1/11,-2

Mathematics
1 answer:
boyakko [2]3 years ago
5 0
least from greatest
-2.1, -2, 1/11, 2
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Which function represents a reflection of f(x) = 5(0.8)^x across the x-axis?
Vikki [24]

Answer: The correct option is second.

Explanation:

The given function is,

f(x)=5(0.8)^x

If we graph a function is f(x), then its coordinates is defined as (x,f(x)).

When the graph of f(x) is reflect across the x-axis, then the the x-coordinate remains the same and the sign of y-coordinate is changed. It means after reflecting across the x-axis,

(x,y)\rightarrow(x,-y)

The given given equation can be written as,

y=5(0.8)^x

To find the equation of the graph after reflection across the x-axis multiply both sides by -1.

-y=-5(0.8)^x

Because f(x)=y and g(x)=-y.

g(x)=-5(0.8)^x

Therefore the second option is correct and the graph of both function is given below.

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How to solve this ? I am not sure pls if you know the answer answer it I really neeed it for marks
xz_007 [3.2K]

Answer:

L.S = R.S ⇒ Proved down

Step-by-step explanation:

Let us revise some rules in trigonometry

  1. sin²α + cos²α = 1
  2. sin2α = 2 sin α cosα
  3. cscα = 1/sinα

To solve the question let us find the simplest form of the right side and the left side, then show that they are equal

∵ L.S = csc2α + 1

→ By using the 3rd rule above

∴ L.S = \frac{1}{sin2\alpha} + 1

→ Change 1 to \frac{sin2\alpha}{sin2\alpha}

∴ L.S = \frac{1}{sin2\alpha} + \frac{sin2\alpha}{sin2\alpha}

→ The denominators are equal, then add the numerators

∴ L.S = \frac{1+sin2\alpha}{sin2\alpha}

∵ R. S = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α

∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα

→ By using the 1st rule above, equate sin²α + cos²α by 1

∴ (sinα + cosα)² = 1 + 2 sinα cosα

→ By using the 2nd rule above, equate 2 sinα cosα by sin2α

∴ (sinα + cosα)² = 1 + sin2α

→ Substitute it in the R.S above

∴ R. S = \frac{1+sin2\alpha}{sin2\alpha}

∵ L.S = R.S

∴ csc 2α + 1 = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

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3 years ago
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