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1 year ago

Which expression is equivalent to 3sqrtx^5y?

Mathematics

1 answer:

slamgirl [31]1 year ago

3 0

Answer:

The correct option is B.

Step-by-step explanation:

The given expression is

$\sqrt[3]{x^5y}$

$\sqrt[3]{x^5y}=(x^5y)^{\frac{1}{3}}$ $[\because \sqrt[n]{x}=x^{\frac{1}{n}}]$

$\sqrt[3]{x^5y}=(x^5)^{\frac{1}{3}}y^{\frac{1}{3}}$ $[\because (mn)^x=m^xn^x]$

$\sqrt[3]{x^5y}=x^{\frac{5}{3}}y^{\frac{1}{3}}$ $[\because (x^m)^n=x^{mn}]$

Therefore option B is correct.

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Which properties are present in a table that represents an exponential function in the form y-b* when b > 1?

Oksana_A [137]

Answer:

<u>Properties that are present are </u>

Property I

Property IV

Step-by-step explanation:

The function given is $y=b^x$ where b > 1

Let's take a function, for example, $y=2^x$

Let's check the conditions:

I. As the x-values increase, the y-values increase.

Let's put some values:

y = 2 ^ 1

y = 2

and

y = 2 ^ 2

y = 4

So this is TRUE.

II. The point (1,0) exists in the table.

Let's put 1 into x and see if it gives us 0

y = 2 ^ 1

y = 2

So this is FALSE.

III. As the x-value increase, the y-value decrease.

We have already seen that as x increase, y also increase in part I.

So this is FALSE.

IV. as the x value decrease the y values decrease approaching a singular value.

THe exponential function of this form NEVER goes to 0 and is NEVER negative. So as x decreases, y also decrease and approached a value (that is 0) but never becomes 0.

This is TRUE.

Option I and Option IV are true.

7 0

2 years ago

Read 2 more answers

Each month for 5 months, sophie makes 2 quilts. how many more quilts does she need to make before she has made 16 quilts?

Zigmanuir [339]

6 quilts are needed

5 0

2 years ago

Read 2 more answers

Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies a

Butoxors [25]

Answer:

$P =0.3998$

Step-by-step explanation:

Let ${\overline{x}}$ be the average of the sample, and the population mean will be $\mu$

We know that:

$\mu = 4095$ gr

Let $\sigma$ be the standard deviation and n the sample size, then we know that the standard error of the sample is:

$E=\frac{\sigma}{\sqrt{n}}$

Where

$\sigma=569$

$n=130$

In this case we are looking for:

$P(|{\displaystyle{\overline{x}}}- \mu|>42)$

This is:

${\displaystyle{\overline{x}}}- \mu>42$ or ${\displaystyle{\overline{x}}}- \mu$

$P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu$

Now we get the z score

$Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z$

$P=P(z>0.8416) + P(z$

Looking at the tables for the standard nominal distribution we get

$P =0.1999+0.1999$

$P =0.3998$

6 0

1 year ago

Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents

Orlov [11]

Step-by-step answer:

The base of the exponential function is 1.29 for 7 days, as in

f(x) = 86*(1.29)^x

The new rate for days can be calculated by dividing x by 7 (where x remains the number of weeks), namely

f(x) = 86*1.29^(x/7)

Using the law of exponents, b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

we simplify by putting b=1.29, a=7 to get

f(x) = 86*(1.29^(1/7))^x

f(x) = 86*(1.037)^x since 1.29^(1/7) evaluates to 1.037

Rounding 1.037 to 1.04 we get a (VERY) approximate function

f(x) = 86 * (1.04^x)

1.04 is very approximate because 1.04^7 is supposed to get back 1.29, but it is actually 1.316, while 1.037^7 gives 1.2896, much closer to 1.29.

7 0

1 year ago

Read 2 more answers

Mick completes a total of 5 assessments for his course. He gets these marks 56,42,47,59,48. Mike needs an average mark of 50 or

Mars2501 [29]

To find the average assessment, add the 5 grades together, then divide by the quantity of grades added together (in this case 5).

=56 + 42 + 47 + 59 + 48
=252

=252 ÷ 5
=50.4 average

ANSWER: Yes, he passes his course with an average grade of 50.4.

Hope this helps! :)

4 0

2 years ago