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3 years ago

Which of the following is the graph of y = –2(2)x + 2?

Mathematics

1 answer:

Dafna11 [192]3 years ago

6 0

Answer: Second graph is the correct option.

Step-by-step explanation:

Since we have given that

$y=-2(2)x+2\\\\y=-4x+2$

We will find the coordinate on y - axis first,

for coordinates on y-axis then x must be equal to 0.

so, put x=0,

$y=-4(0)+2\\\\y=0+2\\\\y=2$

So, according to this, Second graph has satisfied it.

Hence, Second graph is the correct option.

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let sin(θ) =3/5 and tan(y) =12/5 both angels comes from 2 different right trianglesa)find the third side of the two tringles b)

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In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

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$x^2+3^2=5^2$

Solving for x, we have

$\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}$

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

$\tan (y)=\frac{12}{5}$

Describes the following triangle

Using the Pythagorean Theorem again, we have

$5^2+12^2=h^2$

Solving for h, we have

$\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}$

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

$\begin{gathered} \sin (\theta)=\frac{3}{5} \\ \cos (\theta)=\frac{4}{5} \\ \sin (y)=\frac{12}{13} \\ \cos (y)=\frac{5}{13} \end{gathered}$

To calculate the sine and cosine of the sum

$\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}$

We can use the following identities

$\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}$

Using those identities in our problem, we're going to have

$\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}$

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9 months ago

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Answer:

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(I'm sorry if I was wrong)

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