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3 years ago

Julio’s rotation maps point K(–6, 9) to K’(9, 6). Which describes the rotation?

Mathematics

2 answers:

Lilit [14]3 years ago

6 0

Answer:

Step-by-step explanation:

the correct answer is c 90 degrees counter clockwise

11Alexandr11 [23.1K]3 years ago

5 0

90 degrees clockwise

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Without expanding determinants, Prove that

sasho [114]

Answer:

To prove???

Step-by-step explanation:

i have proved this

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3 years ago

Practice multiplying numbers by powers of 10.

Juliette [100K]

Answer:

Multiply (400 + 20 + 3) x 10000

Add 4 zeros to the end of 423

Step-by-step explanation:

423 x 10,000 = Adding the amount of zero’s in 10000 to 423

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3 years ago

Rewrite the expression without the negative exponent. 2x^−4

romanna [79]

Answer:

$\frac{2}{x^4}$

Step-by-step explanation:

The expression is $2x^{-4}$

We use the rule of exponents shown below to change the original to "positive" exponent form:

$a^-b=\frac{1}{a^b}$

Now we can change the original expression given into positive exponent:

$2x^{-4}\\=\frac{2}{x^4}$

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2 years ago

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Suppose ABC added a new point, D. If the coordinates of D were (5, 2), what would be the coordinates of D' have to be if it foll

atroni [7]

Answer:

C the coordinates are just fliped since It was flipped on the y- axis

Step-by-step explanation:

7 0

2 years ago

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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