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tankabanditka [31]

2 years ago

15

The point (Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, StartFraction StartRoot 2 EndRoot Over 2 EndFraction)

is the point at which the terminal ray of angle Theta intersects the unit circle. What are the values for the cosine and cotangent functions for angle Theta? cosine theta = Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, cotangent theta = negative 1 cosine theta = StartFraction StartRoot 2 EndRoot Over 2 EndFraction, cotangent theta = 1 cosine theta = StartFraction StartRoot 2 EndRoot Over 2 EndFraction, cotangent theta = negative one-half cosine theta = Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, cotangent theta = one-half

2 answers:

Ierofanga [76]2 years ago

5 0

Answer:

$\cot(x) = - 1 \: and \: \cos(x) = - \frac{ \sqrt{2} }{2}$

Step-by-step explanation:

The given point is :

$( - \frac{ \sqrt{2} }{2}, \frac{ \sqrt{2} }{2})$

This point is in the second quadrant.

This means:

$\cos(x) = - \frac{ \sqrt{2} }{2}, \sin(x) = \frac{ \sqrt{2} }{2})$

Cotangent is cosine/sine

$\cot(x)=\frac{ \frac{ \sqrt{2} }{2} }{ - \frac{ \sqrt{2} }{2} } = - 1$

Taya2010 [7]2 years ago

4 0

Answer:

A.

Step-by-step explanation:

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Luda [366]

Answer:

B - Spider-Man

Step-by-step explanation:

8 0

2 years ago

An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t

Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61$

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

$Z = \frac{246 - 245}{2.61}$

$Z = 0.38$

$Z = 0.38$ has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0

2 years ago

Solve the following addition and subtraction problems. a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam b. 5 sq.km 95 ha 8,994 sq.m + 11 sq.

kipiarov [429]

As a general rule to solve the problem we are going to transform all values to the lower unit.
a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam
3,000 m 900 m 90 m 19 m + 7,000 m 70 m = 4,009 + 7,070 = 11,079 m
b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m
c. 44 m â€“ 5 dm
44 m - 0.5 m = 43.5 m
d. 73 km 47 hm 2 dam - 11 km 55 hm

73,000 m 4,700 m 20 m - 11,000 m 5,500 m
77,720 m - 16,500 m = 61,220 m

5 0

2 years ago

Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right

Jet001 [13]

Answer:

Option B.

Step-by-step explanation:

It is given that ΔSRQ is a right angle triangle, ∠SRQ is right angle.

RT is altitude on side SQ, ST=9, TQ=16 and SR=x.

In ΔSRQ and ΔSTR,

$m\angle S=m\angle S$ (Reflexive property)

$m\angle R=m\angle T$ (Right angle)

By AA property of similarity,

$\triangle SRQ\sim \triangle STR$

Corresponding parts of similar triangles are proportional.

$\dfrac{SR}{SQ}=\dfrac{ST}{SR}$

Substitute the given values.

$\dfrac{x}{9+16}=\dfrac{9}{x}$

$\dfrac{x}{25}=\dfrac{9}{x}$

On cross multiplication we get

$x^2=25\times 9$

$x^2=225$

Taking square root on both sides.

$x=\sqrt{225}$

$x=15$

The value of x is 15. Therefore, the correct option is B.

7 0

2 years ago

Read 2 more answers

The number of newly reported crime cases in a county in New York State is shown in the accompanying table, where x represents th

Novay_Z [31]

Answer:

Step-by-step explanation:

I use 84+ CE

stat edit, then fill in the #s

then

vars 5

then

2'nd stat plot, on

then, click stat

Click arrow 1 time to the left to get to Calc

then click (4)(LinReg(ax+b))

then click enter 5 times

(y=-25.31428571x+1000.285714

y=-25.3x+1000.3

now, lets use computer:

y=-25.31(543)+1000.3

y=-12743.03

round to the biggest whole number )

this doesn't really work, so I will put 1999, 2000, 2001, 2002, 2003, 2004 instead of 0, 1, 2, 3, 4, 5 and do the same thing

now I get

y=-25.31428571x+51603.54286

y=-25.3x+51603.5

now, lets use computer:

y=-25.3(543)+51603.5

y=37865.6

round to the biggest whole number:

y=37866

so, year 37866

7 0

2 years ago