PLZ HELP! Simplify the radical expression. √64y^10h^5/16y^12h^3

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Answer 1

Answer: Find the attachment for the detailed solution to this question;

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Suppose we toss a coin 4 times. how many different sequences of outcomes are possible if order matters? (for example, hhtt is one outcome and htht is counted as a different outcome)

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Solution: Each coin toss has 2 possible outcomes "Head" and "Tail". So if we flip a coin four times, the number of possible outcomes are:

$2^(4) = 16$ outcomes.

Let H denotes the Head and T denotes the Tail, then the 16 possible outcomes are enumerated below:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

How do i solve this answer ugh please help

Answers

Separate the quadratic functions into two individual function in x which multiply together to equal the original quadratic.
Eg: x^2-3x-10=0 can be written as (x+2)(x-5)=0
The two solutions for x can then be determined by making each bracket equal to zero.
x+2=0 so x=-2x-5=0 so x=5

Solve Sec 5theta/4=2

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$\sec(5\theta)/(4)=2\n (1)/(\cos (5\theta)/(4))=2\n \cos (5\theta)/(4)=(1)/(2)\n (5\theta)/(4)=(\pi)/(3)+2k\pi \vee (5\theta)/(4)=-(\pi)/(3)+2k\pi\n 5\theta=(4\pi)/(3)+8k\pi \vee 5\theta=-(4\pi)/(3)+8k\pi\n \boxed{\theta=(4\pi)/(15)+(8k\pi)/(5) \vee \theta=-(4\pi)/(15)+(8k\pi)/(5)}$

What is the chance of rolling a 4 with a single die three times in a row

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The key thing to remember when you're working a problem like this is that
the cube has no memory ... it doesn't know what happened on the last roll,
so the chance of any particular thing happening on the 2nd roll, or on the
100th roll, is the same as it was on the first roll.

Probability of success = (number of ways to succeed) / (total possibilities)

Number of ways for a 4 to come up . . 1
Total possible results of the roll . . . . . 6

Probability of rolling a 4 . . . . . 1 / 6 .

Probability of rolling two 4s . . . (1/6) x (1/6)

Probability of rolling three 4s . . (1/6) x (1/6) x (1/6) = 1/216 = about 0.46% .

The probability of rolling 4 in one roll is $(1)/(6)$ .
So the probability of rolling 4 three times in a row is $(1)/(6)\cdot(1)/(6)\cdot(1)/(6)=(1)/(216)$

What is 1 1/5 times 2 1/4

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11 divided by 5 is 2.2 and you multiply that by 2 1/4 which is 2.25 in decimal form. The answer is 4.95

Turn 1 1/5 into an improper fracrion witch would be 6/5 and so the same for 2 1/4 witch would be 9/4 then you would multiply an d simplify to get 1 4/5

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