Draw the angle 0=-pi/2 in standard position find the sin and cos

Answers

Answer 1

An angle in standard position has the vertex at the origin and the initial side is on the positive x-axis.

Thus, the initial side of the angle is:

Now, half the circumference measures pi, thus, pi/2 is a quarte of the circumference. As we want to find the angle -pi/2, then we need to rotate the terminal side clockwise:

Find the sine and the cosine.

The sine and the cosine in the unit circle are given by the coordinates as follows:

[tex](\cos\theta,\sin\theta)[/tex]

As can be seen in the given unit circle, the terminal side is located at:

[tex](0,-1)[/tex]

Thus, the values of cosine and sine are:

[tex]\begin{gathered} \cos\theta=0 \\ \sin\theta=-1 \end{gathered}[/tex]

Related Questions

I need help with geometry!

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Basic geometry are formulars and properties of basic shapes like rectangle, square, circle, triangle, and solid shapes like cuboid, cube cylinder etc.

The area, perimeter and volume of solid shape are properties that can be determined from this shape.

Perimeter is the sum of the whole side of the figure. Example the perimeter of a rectangle with 2 length and 2 width can be calculated by adding the whole 2 length and width.

The perimeter of the rectangle above is by adding all the sides.

perimeter = 4 + 4 + 2 + 2 = 12 cm

The area of the figure below is the amount of space of the boundary. The area of the rectangle below is length * width = 4 * 2 = 8 cm squared.

Write this trinomial in factored form. 5a² - 30 - 14

Answers

[tex]5x^2-3x-14[/tex]

replace x with a for this exercise

we use this formula to factor

[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]

where a=5, b=-3 and c=-14

[tex]x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4(5)(-14)}}{2(5)}[/tex][tex]\begin{gathered} x=\frac{3\pm\sqrt[]{9+280}}{10} \\ \\ x=\frac{3\pm\sqrt[]{289}}{10} \\ \\ x=\frac{3\pm17}{10} \end{gathered}[/tex]

we have two roots

[tex]\begin{gathered} x=\frac{3+17}{10} \\ x=2 \end{gathered}[/tex]

and

[tex]\begin{gathered} x=\frac{3-17}{10} \\ \\ x=-\frac{7}{5} \end{gathered}[/tex]

so the simplified equation is

[tex](x-2)(x+\frac{7}{5})[/tex]

now replace x for a

[tex](a-2)(a+\frac{7}{5})[/tex]

I need a little understanding on a one page math assignment

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[tex]\begin{gathered} The\text{ charge of 1 cubic yards at EZ excavation is}\Rightarrow\frac{196}{8}=24.5\text{ \$} \\ The\text{ charge of 1 cubic yards at Happy service is}\Rightarrow25\text{ \$} \\ (a)\text{ For 40 cubic yards is,} \\ At\text{ EZ excavation}\Rightarrow40\times24.5=\text{ \$980} \\ At\text{ Happy service}\Rightarrow40\times25=\text{ \$1000} \end{gathered}[/tex]

If I take a 45 min. break at 2:15pm what time do I come back?

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The break time is 2:15 pm.

The time interval for break is 45 min.

Determine the time at which interval ends.

[tex]\begin{gathered} 2\colon15+00.45=2\colon60 \\ =3\colon00 \end{gathered}[/tex]

So break ends (individual come back) at 3:00.

The function y=f(x) is graphed below. Plot a line segment connecting the points on ff where x=-1 and x=0. Use the line segment to determine the average rate of change of the function f(x) on the interval −1≤x≤0

Answers

Answer:

Aveage Rate of cCanege = 40

Explanation:

The line segment is drawn in the function below:

Using the line segment:

[tex]\begin{gathered} \Delta x=0-(-1)=1 \\ \Delta y=40-0=40 \end{gathered}[/tex]

Therefore, the average rate of change will be:

[tex]\text{ Average Rate of Change}=\frac{\Delta y}{\Delta x}=\frac{40}{1}=40[/tex]

The average rate of change is 40.

y varies directly as x, y = 7 when x = 21. Determine x when y = 5.

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y varies directly as x, y = 7 when x = 21. Determine x when y = 5.

Step 1

Let

y varies directly as x, it is y depends on x, in math terms

f(x)=y

y = 7 when x = 21

f(21)=7

Determine x when y = 5. f(?)=5

Step 2

there is a proportion, this must be equal, make a rule of three to find the value

x y

[tex]\begin{gathered} 21\leftrightarrow7 \\ x\text{ }\leftrightarrow5 \\ \text{the relation is} \\ \frac{21}{7}=\frac{x}{5} \\ \text{solve for x} \\ x=\frac{21\cdot5}{7} \\ x=\frac{105}{7} \\ x=15 \end{gathered}[/tex]

so , when y=5, x=15

The average score for games played in the NFL is 22 and the standard deviation is 9.3 points. 41 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.

a. What is the distribution of ¯x x¯
? ¯xx¯ ~ N( , )
b. What is the distribution of ∑x ? ∑x ~ N ( , )
c. P( ¯x > 19.8214) =
d. Find the 60th percentile for the mean score for this sample size.
e. P(20.6214 < x¯< 23.2262) =
f. Q1 for the x¯distribution =
g. P( ∑x > 829.0774) =

For part c) and e), Is the assumption of normal necessary? NoYes

Answers

Using the normal distribution and the central limit theorem, it is found that:

a) The distribution is: x¯ ~ N(22, 1.45).

b) The distribution is: ∑x ~ N(902, 59.55).

c) P( ¯x > 19.8214) = 0.9332 = 93.32%.

d) The 60th percentile for the mean score for this sample size is of 22.37 points a game.

e) P(20.6214 < x¯< 23.2262) = 0.6312 = 63.12%.

f) Q1 for the x¯distribution = 21 points a game.

g) P( ∑x > 829.0774) = 0.8888 = 88.88%.

Assumption of normality is not necessary, as the sample sizes are greater than 30.

Normal Probability Distribution

The z-score of a measure X of a variable that has mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure X is above or below the mean of the distribution, depending if the z-score is positive or negative.From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Also by the Central Limit Theorem, for the sum of n instances of a variable, the mean is of [tex]\n\mu[/tex] and the standard deviation is of [tex]\sigma\sqrt{n}[/tex].Finally, by the Central Limit Theorem, assumption of normality is only necessary when the sample size is less than 30.

For a single game, the mean and the standard deviation of the number of points scored are given as follows:

[tex]\mu = 22, \sigma = 9.3[/tex]

For the average of 41 games, the standard error is:

[tex]s = \frac{9.3}{\sqrt{41}} = 1.45[/tex]

Hence the distribution is: x¯ ~ N(22, 1.45).

For the sum of the 41 games, the mean and the standard error are given as follows:

41 x 22 = 902.[tex]s = 9.3\sqrt{41} = 59.55[/tex].

Hence the distribution is: ∑x ~ N(902, 59.55).

In item c, the probability is one subtracted by the p-value of Z when X = 19.8214, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem: