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3 years ago

Provide a counterexample: sinx = 1/sinx

1 answer:

8 0

Sin x = 1/sin x
$sin^{2} x=1 \\ sin x= \pm1 \\ sinx=1 \ or \ sinx=-1 \\ x= sin^{-1} 1 \ or \ x= sin^{-1} (-1) \\ x= \frac{ \pi }{2} \ or \ x= \frac{3 \pi }{2}$

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An engineer is building a model of a vehicle, using a 43 : 2 scale.The actual dimensions of the vehicle are given in the table.

gayaneshka [121]

The correct answer is (E) 8.9

Hope you were able to pass the test <3

3 0

3 years ago

Elyse wants to buy a new softball glove that costs $45. She has $15 and plans to save 5 each week. How many weeks will it take h

almond37 [142]

Answer:

6 weeks :)

Step-by-step explanation:

it will take her 6 weeks

45-15 is 30.

30/5 is 6

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3 years ago

Read 2 more answers

Please help! I am confused

Levart [38]

I would say the 60 + 20 represents how the shirts cost the same amount. Then I would say the x either represents the cost before tax or how to find the cost of the shirts.

I hope this helps!
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3 years ago

Read 2 more answers

A fisherman can row upstream at 1 mph and downstream at 4 mph he started rowing upstream until he got tired and then rowed downs

Morgarella [4.7K]

Answer:the total trip, and I should be right, was 7 miles. He rowed up for 3 hours, and back for 1.

8 0

4 years ago

Find the probability of being dealt a full house. (Round your answer to six decimal places.)?

vesna_86 [32]

I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.

A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.

Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.

The number of ways of drawing 2 aces and 3 2s is

$\dbinom42\dbinom43=24$

and the number of ways of drawing 3 aces and 2 2s is the same,

$\dbinom43\dbinom42=24$

so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.

Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is

$2\dbinom{13}2\dbinom42\dbinom43=3744$

The total number of hands that can be drawn is

$\dbinom{52}5=2,598,960$

Then the probability of getting a full house is

$\dfrac{2\binom{13}2\binom42\binom43}{\binom{52}5}=\dfrac6{4165}\approx\boxed{0.001441}$

4 0

3 years ago