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Using the unit circle, determine the value of sin 450°

Answers

Answer 1
Answer:

Answer:

1

Step-by-step explanation:

To find the value of sin 450 degrees using the unit circle, represent 450° in the form (1 × 360°) + 90° [∵ 450°>360°] ∵ sine is a periodic function, sin 450° = sin 90°.

Rotate ‘r’ anticlockwise to form a 90° or 450° angle with the positive x-axis.

The sin of 450 degrees equals the y-coordinate(1) of the point of intersection (0, 1) of unit circle and r.

Hence the value of sin 450° = y = 1
Answer 2
Answer:

Answer:

To find the value of sin 450 degrees using the unit circle, represent 450° in the form (1 × 360°) + 90° [∵ 450°>360°] ∵ sine is a periodic function, sin 450° = sin 90°.

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Find the scale factor of the line segment dilation. AB: endpoints (-6, -3) and (-3,-9) to A'B': endpoints at (-2, -1) and (-1, -3). A) -1/3 B) 1/3 C) 3D) -3

Answers

we know that the endpoints of AB are

\begin{gathered} (x_1,y_1)=\mleft(-6,-3\mright) \n (x_2,y_2)=(-3,-9) \end{gathered}

and the distance formula is given by

d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}

By substituying these points, we have that

d=\sqrt[]{(-3-(-6))^2+(-9-(-3))^2}

which is equal to

\begin{gathered} d=\sqrt[]{(-3+6)^2+(-9+3)^2} \n d=\sqrt[]{3^2+(-6)^2} \end{gathered}

then

\begin{gathered} d=\sqrt[]{9+36} \n d=\sqrt[]{45} \n d=\sqrt[]{9\cdot5} \n d=\sqrt[]{9}\cdot\sqrt[]{5} \n d=3\sqrt[]{5}\ldots..(A) \end{gathered}

On the other hand, if

\begin{gathered} (x_1,y_1)=(-2,-1) \n (x_2,y_2)=(-1,-3) \end{gathered}

similarly to the previous case, the distance between the endpoint for A'B' is

d=\sqrt[]{(-1-(-2))^2+(-3-(-1))^2}

which is equal to

\begin{gathered} d=\sqrt[]{(-1+2)^2+(-3+1)^2} \n d=\sqrt[]{1^2+(-2)^2} \n d=\sqrt[]{1+4} \n d=\sqrt[]{5}\ldots..(B) \end{gathered}

Now, by comparing equation A and equation B, we can see that, the scale factor is 1/3.

Then, the answer is B.

49, 34, and 48 students are selected from the Sophomore, Junior, and Senior classes with 496, 348, and 481 students respectively. Identify which type of sampling is used and why?A) Convenience
B) Cluster
C) Random
D) Systematic
E) Stratified

Answers

Answer:

Stratified sampling

Step-by-step explanation:

In the question, students are selected from subgroups of Sophomore, Junior, and Senior classes.

We are told that 49, 34, and 48 students are selected from the Sophomore, Junior, and Senior classes with 496, 348, and 481 students respectively.

This means that the sample numbers of 49, 34 and 48 students were selected in proportion to the subgroup sizes of 496, 348, and 481 students respectively.

Thus, due to the fact that subgroups were selected & that sample number of students were also selected in proportion to their respective subgroup sizes, this is therefore a stratified sampling.

Final answer:

The type of sampling used in this scenario is stratified sampling. The type of sampling in this case is E) Stratified sampling.

Explanation:

The type of sampling used in this scenario is stratified sampling. Stratified sampling is a method where the population is divided into different groups or strata, and then a sample is randomly selected from each stratum. In this case, the sophomore, junior, and senior classes are the different strata, and the sample sizes are proportional to the size of each stratum.

To further explain, let's consider the sophomore class with 496 students. If we are selecting a sample size of 49 from this class, it means that approximately 10% of the students (49/496 = 0.098) will be selected. Similarly, for the junior class with 348 students, approximately 10% (34/348 = 0.098) of the students will be selected. The same applies to the senior class with 481 students and a sample size of 48 students.

In stratified sampling, the goal is to ensure that the sample represents the characteristics of each stratum in proportion to their size in the population. By sampling from each of the classes, we can obtain a representation of the entire student population.

Learn more about Stratified Sampling here:

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Fill in the green box.

Answers

Answer:

y=6

solution,

SimilarRighttriangles:

(c)/(h) = (h)/(d) \n {h}^(2) = cd \n {y}^(2) = 4 * 9 \n {y = 36 }^(2) \n y = \sqrt{ {(6)}^(2) } \n y = 6

Hopethis helps..

Goodluck on your assignment..

Answer:whats the measure of x i really need help

Step-by-step explanation:

Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 * 500 picture elements (pixels), where each pixel can take on one of 32 intensity values. Assume that 30 pictures are sent per second. (This digital source is roughly equivalent to broadcast TV standards that have been adopted.) Find the source rate R (bps).

Answers

Answer:

36,000,000 bps

Step-by-step explanation:

Matrix of 480 by 500 = 480 X 500 = 240,000 pixels

Intensity values per pixel = 32. This simply means the color variations in this TV.

The bit per pixel for an intensity value of 32 is 5 (That's 2 raised to the power of 5)

Pictures sent per second = 30

Source rate R (bps) = Total number of pixels in the matrix X bit per pixel X Number of pictures sent per second.

Source rate R (bps) = 240,000 X 5 X 30 = 36,000,000 bps

Final answer:

The source rate for the digitized TV picture is 230,400,000 bps.

Explanation:

The source rate can be calculated by multiplying the number of pixels per picture element by the number of intensity values per pixel and by the number of pictures sent per second.

So, the source rate R (bps) is given by:

R = (480 * 500) * 32 * 30

Calculating this, we find that the source rate is 230,400,000 bps.

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A smoothie recipe calls for 3 cups of soy milk, 2 frozen bananas and, 1 tablespoon of chocolate syrup. Write a sentence that uses a ratio to describe this recipe.

Answers

Answer:

3:2:1

Step-by-step explanation:

30 + (-87) – 140 + 8A

Answers

Simplify the expression.

−197+8A, Hope this helps, Have a wonderful day!~
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