Need help!! don't know what I'm doing

Mathematics

2 answers:

Anarel [89]3 years ago

5 0

Answer:

Its 9 :)

Step-by-step explanation:

nignag [31]3 years ago

3 0

The answer would be 29 !!

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Let (6,-5) be a point on the terminal side of θ. Find the exact values of sin θ, sec θ, and tan θ.

makvit [3.9K]

Answer:

sinθ = -5/√61

secθ = √61/6

tanθ = -5/6

Step-by-step explanation:

From the given coordinate (6, -5), x = 6 and y = -5. This shows that the point lies in the 4th quadrant. In the fourth quadrant, only cos θ is positive, both sin θ and tan θ are negative.

Let us get the value of the radius 'r' first before calculating the trigonometry identities.

Using the Pythagoras theorem;

$r^2 = x^2 + y^2\\r^2 = 6^2 + (-5)^2\\r^2 = 36+25\\r^2 = 61\\r = \sqrt{61}$

Using SOH, CAH, TOA to get the trigonometry identities;

Given x = 6, y =5 and r = √61

$sin \theta = opp/hyp = y/r\\sin \theta = 5/\sqrt{61} \\$

Since sin θ, is negative in the fourth quadrant, $sin \theta = -5/\sqrt{61} \\$

$cos \theta = adj/hyp = x/r\\cos \theta = 6/\sqrt{61} \\$

$sec \theta = \frac{1}{cos \theta} \\sec \theta = \frac{1}{6/\sqrt{61} } \\sec \theta = \frac{\sqrt{61} }{6}$

For tanθ:

$tan \theta = \frac{opp}{adj} = \frac{y}{x} \\tan \theta = \frac{5}{6}\\$

Since tan θ, is negative in the fourth quadrant, $tan \theta = -5/6$

7 0

4 years ago

"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d

dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$