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12345 [234]
2 years ago
9

Which expression is equivalent to (x^6y^8)^3\x^2y^2

Mathematics
2 answers:
BARSIC [14]2 years ago
8 0

Answer: x^{16}\ y^{22}

Step-by-step explanation:

The given expression : \dfrac{(x^6y^8)^3}{x^2y^2}

Using identity , (a^m)^n=a^{mn} , we have

{(x^6y^8)^3=x^{6\times3}\ y^{8\times3}\\\\=x^{18}\ y^{24}

Now, \dfrac{(x^6y^8)^3}{x^2y^2}=\dfrac{x^{18}\ y^{24}}{x^2\ y^2}

( its also an equivalent expression to given expression.)

Using identity , \dfrac{a^n}{a^m}=a^{n-m} , we have

\dfrac{x^{18}\ y^{24}}{x^2\ y^2}=x^{18-2}\ y^{24-2}\\\\=x^{16}\ y^{22}

Hence, the expression is equivalent to given expression :

x^{16}\ y^{22}

Damm [24]2 years ago
5 0

Answer:

\large\boxed{\dfrac{(x^6y^8)^3}{x^2y^2}=x^{16}y^{22}}

Step-by-step explanation:

\dfrac{(x^6y^8)^3}{x^2y^2}\qquad\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=\dfrac{(x^6)^3(y^8)^3}{x^2y^2}=\dfrac{x^{(6)(3)}y^{(8)(3)}}{x^2y^2}=\dfrac{x^{18}y^{24}}{x^2y^2}\qquad\text{use}\ \dfrac{a^m}{a^n}=a^{m-n}\\\\=x^{18-2}y^{24-2}=x^{16}y^{22}

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Robert has $50 to spend on his utility bills each month. The basic monthly charge is 23.77 electricity costs 0.1117 for each kil
Lilit [14]

Answer:

<em>The maximum number of kilowatt-hours is 235</em>

Step-by-step explanation:

<u>Inequalities</u>

Robert's monthly utility budget is represented by the inequality:

0.1116x + 23.77 < 50

Where x is the number of kilowatts of electricity used.

We are required to find the maximum number of kilowatts-hours used without going over the monthly budget. Solve the above inequality:

0.1116x + 23.77 < 50

Subtracting 23.77:

0.1116x < 50 - 23.77

0.1116x < 26.23

Dividing by 0.1116:

x < 26.23/0.1116

x < 235

The maximum number of kilowatt-hours is 235

7 0
1 year ago
A jet flying at 123 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is 2
aev [14]

Answer:

The magnitude of the necessary lifting force is 7.95N

Step-by-step explanation:

Force = mv^2/r

mass (m) = 2.00105kg, velocity (v) = 123m/s, radius (r) = 3810m

Force = 2.00105×123^2/3810 = 30273.89/3810 = 7.95N

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1 year ago
At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima
Firdavs [7]
\mathbf r(t)=\langle x(t),y(t),z(t)\rangle=\langle\sin t,\cos t,t\rangle

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7 0
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ΔABC is an equilateral triangle. m∠A = (3x - 12)°. Solve for x. A) 14 B) 19 C) 20 D) 24
uranmaximum [27]

Answer:

c is the answer

Step-by-step explanation:

hope it is correct tho

8 0
1 year ago
Read 2 more answers
An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
arlik [135]

Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 15+15-2=28  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
1 year ago
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