Chicago ave is parallel to Ontario Street. What is the relationship between angles 2 and 3?

Given △DEF, which is not equal to cos(F)? sin(F). sin(D). tan(F). cos(D)

valentina_108 [34]

Answer: The correct option is tan(F).

Explanation:

In the given figure the triangle is a right angle triangle because $\angle E=90^{\circ}$ .

It is also given that $DE=EF=5$ and $DF=5\sqrt{2}$ .

Since it is an isosceles right angle triangle, therefore the value of perpendicular and base is same for both angles D and F, which is 5.

$\cos (F)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (F)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (D)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(D) is $\frac{1}{\sqrt{2}}$ .

$\tan (F)=\frac{Perpendicular}{Base}=\frac{5}{5} =1$

The value of tan(F) is 1. Which is not equal to $\frac{1}{\sqrt{2}}$ .

$\cos (D)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(D) is $\frac{1}{\sqrt{2}}$ .

Therefore, the value of tan(F) is not equal to the value of cos(F), so the correct option is third.

3 0

2 years ago

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A, B, and C are polynomials, where A = n, B = 2n + 6, and C = n2 – 1. What is AB – C in simplest form? A=–n2 + 3n + 5 B=n2 + 6n

slava [35]

Answer: B: n^2+6n+1

Step-by-step explanation:

A=n

B=2n+6

C=n^2-1

AB-C

n(2n+6)-n^2-1

2n^2+6n-n^2+1

n^2+6n+1

4 0

2 years ago

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David drives 168 miles a week for work. He fills his gas tank twice a week. Each fill up is for 12 gallons. Assuming he uses all

hammer [34]

If you knew how to divide, you could answer this problem.

3 0

2 years ago

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You and your friend are selling tickets to a charity event. You sell 11 adult tickets and 8 student tickets for $158. Your frien

Crazy boy [7]

{11X+8Y=158
{5X+17Y=152

{55X+40Y=790
-
{55X+187Y=1672

147Y=882
Y=6
Cost of the ticket for student is $6

3 0

2 years ago

Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes

dimaraw [331]

Answer:

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Step-by-step explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let $\mu$ = population mean length of the songs

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis, $H_A$ : $\mu$ > 210 seconds {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is One-sample z-test statistics because we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean length of songs = 252.5 seconds

$\sigma$ = population standard deviation = 54.5 seconds

n = sample of current hit songs = 40

So, the test statistics = $\frac{252.5-210}{\frac{54.5}{\sqrt{40} } }$

= 4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

4 0

2 years ago