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3 years ago

Thanks for the help!

Mathematics

1 answer:

NikAS [45]3 years ago

3 0

Answer:

Step-by-step explanation:

So we have the piecewise function:

$f(x)= \begin{cases}\sqrt{2x} & \text{if }x$

And we want to find f(8).

Since our input value is 8, choose the equation that fits our input.

The first equation demand x to be less than 3. 8 is not less than 3, so we won't use that.

The second equation demand x to e greater than or equal to 3 and less than 8. 8 is <em>not</em> less than 8. So, we won't use that.

The third equation demands x to be greater than or equal to 8. 8 is greater than or equal to 8, so we'll use the third equation.

So:

$f(x)=42\text{ if }x\geq 8$

$f(8)=42$

There're nothing more to do, we're done :)

The answer is A.

Edit: Improved Format

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Which term can be added to the list so that greatest common factor of the three terms is 12h^3? 36h^3 12h^6. 6h^3. 12h^2. 30h^4.

Juliette [100K]

Answer:

Last Option $48h^{5}$ .

Step-by-step explanation:

In this question we will do the factors of given two terms then we will do the common factors of all the terms given in answer to match with.

Common factors of 36h³ = 1×2×2×3×3×h×h×h

Common factors of $12h^{6}$ = 1×2×2×3×h×h×h×h×h×h

In these two terms greatest common factor Of these two terms is = 1×2×2×3×h×h×h = 12h³

Therefore the third term will be the number which has the greatest common factor = 12h³

So the given terms are

6h³ = 1×2×3×h×h×h

12h² = 1×2×2×3×h×h

$30h^{4}$ = 1×2×3×5×h×h×h×h

$48h^{5}$ = 1×2×2×2×2×3×h×h×h×h×h

Therefore the greatest common factor of the term which matches with 12h³ is $48h^{5}$

3 0

2 years ago

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$