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3 years ago

Write the trigonometric expression Cos (arcsin (u)) As an algebraic expression in u

1 answer:

5 0

When we use arcsine, we are finding the angle while giving the trigonometric ratio.

Arcsin(u) = theta can be rewritten as:

sin(theta) = u

Sine is opposite over hypotenuse, so u/1 means that the side opposite to theta (the y value) is u, and the hypotenuse is 1.

We can use Pythagorean Theorem to find the adjacent (x value).

1^2 - u^2 = x^2

x = sqrt(1-u^2)

Back to the original question, we are trying to find cos(arcsin(u)). We just solved all the sides for our triangle using arcsin(u). Now we need to do cos(u).

Cosine is adjacent over hypotenuse.

So our answer is sqrt(1-u^2)/1

Or just sqrt(1-u^2)

You might be interested in

(09.02)

Ivenika [448]

Answer:

4th option

Step-by-step explanation:

x² - 2x - 3 = 0 ( add 3 to both sides )

x² - 2x = 3

using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 1)x + 1 = 3 + 1

(x - 1)² = 4

4 0

2 years ago

Maggie's brother is 11 years younger than three times her age. The sum of their ages is 41.

brilliants [131]

Answer:

Maggie is 13 years old.

Step-by-step explanation:

The explanation is attached.

3 0

2 years ago

Find the volume r:7 h:7

sweet-ann [11.9K]

Answer:

V = 1077.566 units²

Step-by-step explanation:

Assuming this is a cylinder, the formula is (area of base) × (height)

or (πr²)(h), simply plug-in the given values:

= (π(7)²)(7)

= (153.938)(7)

V = 1077.566 units²

3 0

3 years ago

A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

DochEvi [55]

Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

$Z = \frac{X - \mu}{\sigma}$

$1.175 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.175*4.7$

$X = 27.0225$

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.04*4.7$

$X = 26.388$

The lowest score that would meet the colleges requirement would be decreased to 26.388.

6 0

3 years ago

I desperately need this question answered. See attached image thanks so much

Dahasolnce [82]

For this case we must follow the steps below:

step 1:

We place each of the given points on a coordinate axis

Step 2:

We join the AC points (represented by the orange line)

We join the BD points (represented by the blue line)

It is observed that the resulting figure after placing the 4 points on a coordinate axis, turns out to be a rhombus.

In addition, the blue and orange lines turn out to be perpendicular, that is, they have an angle of 90 degrees between them. This can be verified by finding the slopes of each of the two straight lines (blue and orange), which must be opposite reciprocal, that is, they comply: $m1 * m2 = -1$

In this case, the slope of the orange line is $m1 = 1$ and that of the blue line is $m2 = -1$

Then $m1 * m2 = 1 * -1 = -1$ , it is verified that they are perpendicular.

Thus, the conclusion is that ABCD is a rhombus and AC is perpendicular to BD.

Answer:

See attached image

Option D

6 0

3 years ago