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Dmitriy789 [7]
3 years ago
10

Big money butts how to do.​

Mathematics
1 answer:
Elenna [48]3 years ago
7 0

Answer:

I have no idea what this means lol

Note:

Please mark as Brainliest! <3

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Can you guys answer these questions please i have allot of work and im trying to do multiple things i know its allot of question
mr_godi [17]

Answer:

um

Step-by-step explanation:

whats the question?

4 0
3 years ago
Which expression is equivalent to log_5(x/4)^2?
viktelen [127]

For this case we must find an expression equivalent to:

log_ {5} (\frac {x} {4}) ^ 2

So:

We expanded log_ {5} ((\frac {x} {4}) ^ 2)by moving 2 out of the logarithm:

2log_ {5} (\frac {x} {4})

By definition of logarithm properties we have to:

The logarithm of a product is equal to the sum of the logarithms of each factor:

log (xy) = log (x) + log (y)

The logarithm of a division is equal to the difference of logarithms of the numerator and denominator.

log (\frac {x} {y}) = log (x) -log (y)

Then, rewriting the expression:

2 (log_ {5} (x) -log_ {5} (4))

We apply distributive property:

2log_ {5} (x) -2log_ {5} (4)

Answer:

An equivalent expression is:

2log_ {5} (x) -2log_ {5} (4)

3 0
3 years ago
Researchers for a company that manufactures batteries want to test the hypothesis that the mean battery life of their new batter
guapka [62]

(b) one-sample t-test for a population mean

ur welcome :D

8 0
3 years ago
A number is divided by 18 the result is 20 what is the number<br>​
Basile [38]

Answer:

360

Step-by-step explanation:

You can solve this by doing 18 times 20 is 360: and then dividing 360 by 18, results in 20.

3 0
3 years ago
Read 2 more answers
Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\
motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in \sqrt{7}, 1 - \sqrt{6} and -3. The last root gives us the factor (x+3). Hence, our polynomial is

P(x) =(x+3)q(x)

where q is a polynomial with rational coefficients and roots \sqrt{7} and 1 - \sqrt{6}. The root \sqrt{7} gives us a factor x-\sqrt{7}, but in order to obtain rational coefficients we must consider the factor x^2-7.

An analogue idea works with 1 - \sqrt{6}. For convenience write  x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}. This gives the factor (x-1)^2-6. Hence,

P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105

Notice that P(-1)=24. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type x-\sqrt7 will introduce in the expression, we need to multiply by its conjugate x+\sqrt7. Hence, we will obtain x^2-7 that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0
3 years ago
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