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snow_tiger [21]

3 years ago

14

Which graph does NOT represent a function?

2 answers:

sergeinik [125]3 years ago

7 0

Answer:

the second one

i just did it

Step-by-step explanation:

Anna007 [38]3 years ago

4 0

Answer:

the circle but im in between v to

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2.943÷2.7 any help???

Zinaida [17]

The answer will be 1.09. Hope this will help you!

6 0

4 years ago

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers

Vera_Pavlovna [14]

Answer:

arc AC = 111°

Step-by-step explanation:

∠APC is a central angle, so arc AC is equal to the measure of ∠APC.

∠APC and ∠BPC are supplementary. So, m∠APC + m∠BPC = 180

m∠APC + 69 = 180

m∠APC = 111

arc AC = 111°

8 0

3 years ago

Solve Highschool Pre cal

Bezzdna [24]

Option C:

$\ln 64 y^{2}$

Solution:

Given expression: 2 ln 8 + 2 ln y

Applying log rule $a \log b=\log b^{a}$ in the above expression.

⇒ $2 \ln 8+2 \ln y=\ln 8^{2}+\ln y^{2}$

Applying log rule $\log a+\log b=\log (a b)$ in the above expression.

⇒ $\ln 8^{2}+\ln y^{2}=\ln \left(8^{2} y^{2}\right)$

We know that $8^{2}=64$

⇒ $\ln \left(8^{2} y^{2}\right)=\ln 64 y^{2}$

Hence, the expression in single natural logarithm is $\ln 64 y^{2}$ .

6 0

3 years ago

HELP ME PLEASE!! I NEED HELP ASAP

jok3333 [9.3K]

Answer:

5

Step-by-step explanation:

5/25 = 1/5

5 is 1/5 of 25

8 0

3 years ago

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