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3 years ago

Which is the graph of the function f(x) = Negative StartRoot x EndRoot?

Mathematics

2 answers:

emmasim [6.3K]3 years ago

6 0

Answer:

The answer is the third graph or c

Step-by-step explanation:

I just did the test and the other guy is correct to.

Maksim231197 [3]3 years ago

6 0

Answer:

If you are on edg 2021 then it is the graph with the line going down from 0 to the bottom left square of the graph

Step-by-step explanation:

So it is the first graph.

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4 years ago

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3 years ago

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5 0

3 years ago

Please help, trig is a real problem for me

Andreyy89

Answer:

2. cotФ = $-\frac{4}{3}$

3. sec²Ф - tan²Ф = 1

4. $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = cos²∝

5. The value of sin x is $\frac{1}{2}$

6. cos 15° = $\frac{\sqrt{6}+\sqrt{2}}{4}$

7. tan(α + β) = $-\frac{63}{16}$

8. sin(π - Ф) = sin Ф

9. cos 2Ф = $-\frac{1}{2}$

10. sin 2Ф = 0.96

Step-by-step explanation:

∵ cos Ф = $-\frac{4}{5}$

∵ 90° < Ф < 180°

- That means Ф lies on the 2nd quadrant

∴ sin Ф is a positive value

∵ sin²Ф + cos²Ф = 1

∴ sin²Ф + ( $-\frac{4}{5}$ )² = 1

∴ sin²Ф + $\frac{16}{25}$ = 1

- Subtract from both sides $\frac{16}{25}$

∴ sin²Ф = $\frac{9}{25}$

- Take √ for both sides

∴ sinФ = $\frac{3}{5}$

∵ cotФ = cosФ ÷ sinФ

∴ cotФ = $-\frac{4}{5}$ ÷ $\frac{3}{5}$

∴ cotФ = $-\frac{4}{3}$

The expression is sec²Ф - tan²Ф

∵ tan²Ф = sec²Ф - 1

- Substitute tan²Ф by the right hand side in the expression

∴ sec²Ф - tan²Ф = sec²Ф - (sec²Ф - 1)

∴ sec²Ф - tan²Ф = sec²Ф - sec²Ф + 1

- Simplify the right hand side

∴ sec²Ф - tan²Ф = 1

The expression is $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$

∵ The numerator is sin²∝ + cos²∝

∵ sin²∝ + cos²∝ = 1

∴ The numerator = 1

∵ The denominator is tan²∝ + 1

∵ tan²∝ = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ + 1

- Change 1 to fraction $\frac{cos^{2}\alpha}{cos^{2}\alpha}$

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ + $\frac{cos^{2}\alpha}{cos^{2}\alpha}$

- Add the two fractions

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha+cos^{2}\alpha }{cos^{2}\alpha}$

∵ sin²∝ + cos²∝ = 1

∴ tan²∝ + 1 = $\frac{1}{cos^{2}\alpha}$

∴ The denominator = $\frac{1}{cos^{2}\alpha}$

∴ $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = $\frac{1}{\frac{1}{cos^{2}\alpha}}$

- Remember denominator the denominator will be a numerator

∴ $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = cos²∝

∵ tan x cos x = $\frac{1}{2}$

∵ tan x = $\frac{sin(x)}{cos(x)}$

∴ $\frac{sin(x)}{cos(x)}$ × cos x = $\frac{1}{2}$

- Simplify it by canceling cos x up with cos x down

∴ sin x = $\frac{1}{2}$

∴ The value of sin x is $\frac{1}{2}$

∵ cos(Ф - ∝) = cosФ cos∝ + sinФ sin∝

∵ 45 - 30 = 15

∴ cos 15° = cos(45 - 30)°

- Use the rule above to find the exact value

∵ cos(45 - 30)° = cos 45° cos 30° + sin 45° sin 30°

∵ cos 45° = $\frac{\sqrt{2}}{2}$ and sin 45° =

∵ cos 30° = $\frac{\sqrt{3}}{2}$ and sin 30° = $\frac{1}{2}$

∴ cos(45 - 30)° = $\frac{\sqrt{2}}{2}$ × $\frac{\sqrt{3}}{2}$ +

∴ cos(45 - 30)° = $\frac{\sqrt{6}}{4}$ + $\frac{\sqrt{2}}{4}$ = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∴ cos 15° = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∵ tan(α + β) = $\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$

∵ cos α = $\frac{5}{13}$ and 0° < α < 90°

- That means α is in the 1st quadrant, then all trigonometry

ratios are positive

∵ sin²α + cos²α = 1

∴ sin²α + ( $\frac{5}{13}$ )² = 1

∴ sin²α + $\frac{25}{169}$ = 1

- Subtract $\frac{25}{169}$ from both sides

∴ sin²α = $\frac{144}{169}$

- Take √ for both sides

∴ sin α = $\frac{12}{13}$

∵ tan α = sin α ÷ cos α

∴ tan α = $\frac{12}{13}$ ÷ $\frac{5}{13}$

∴ tan α = $\frac{12}{5}$

∵ sin β = $\frac{3}{5}$ and 0° < β < 90°

∵ sin²β + cos²β = 1

∴ ( $\frac{3}{5}$ )² + cos²β = 1

∴ $\frac{9}{25}$ + cos²β = 1

- Subtract $\frac{9}{25}$ from both sides

∴ cos²β = $\frac{16}{25}$

- Take √ for both sides

∴ cos β = $\frac{4}{5}$

∵ tan β = sin β ÷ cos β

∴ tan β = $\frac{3}{5}$ ÷ $\frac{4}{5}$

∴ tan β = $\frac{3}{4}$

- Substitute the values of tan α and tan β in the tan (α + β)

∵ tan(α + β) = $\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$

∴ tan(α + β) = $\frac{\frac{12}{5}+\frac{3}{4}}{1-(\frac{12}{5})(\frac{3}{4})}$

∴ tan(α + β) = $-\frac{63}{16}$

∵ sin(α - β) = sin α cos β - cos α sin β

∴ sin(π - Ф) = sin π cos Ф - cos π sin Ф

∵ sin π = 0 and cos π = -1

∴ sin(π - Ф) = (0) × cos Ф - (-1) × sin Ф

∴ sin(π - Ф) = 0 + sin Ф

∴ sin(π - Ф) = sin Ф

∵ cos 2Ф = 2 cos²Ф - 1

∵ cos Ф = $\frac{1}{2}$

∴ cos 2Ф = 2 ( $\frac{1}{2}$ )² - 1

∴ cos 2Ф = 2 × $\frac{1}{4}$ - 1

∴ cos 2Ф = $\frac{1}{2}$ - 1

∴ cos 2Ф = $-\frac{1}{2}$

10.

∵ sin 2Ф = 2 sinФ cosФ

∵ cosФ = 0.6 and 0° < Ф < 90°

- That means Ф is in the 1st quadrant and all its trigonometry

ratios are positive

∵ sin²Ф + cos²Ф = 1

∴ sin²Ф + (0.6)² = 1

∴ sin²Ф + 0.36 = 1

- Subtract 0.36 from both sides

∴ sin²Ф = 0.64

- Take √ for both sides

∴ sinФ = 0.8

- Substitute the values of sinФ and cosФ in the rule above

∴ sin 2Ф = 2(0.8)(0.6)

∴ sin 2Ф = 0.96

4 0

4 years ago