Find all zeros for (2x-1)(x-5)

The length of a rectangular garden is 7x + 3 feet. The width of the garden is

TiliK225 [7]

Answer:

42xsquared-3x-9

Step-by-step explanation:

5 0

3 years ago

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

3 0

3 years ago

The deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipel

otez555 [7]

Answer:

The calculated value of t= -0.21 does not lie in the critical region t= ± 2.120 Therefore we accept our null hypothesis that there's no difference between strength (psi) of liner specimens when a certain fusion process was used and when this process was not used at the 0.05 significance level

Step-by-step explanation:

The given data is

No Fusion :: 2748 2700 2655 2822 2511 3149 3257 3213 3220 2753

Yes Fusion :: 3027 3356 3359 3297 3125 2910 2889 2902

The calculated Means are

x1`= 2902.8

X2`= 3108.125

The standard deviations are

s1= 277.265

s2= 205.87

We are using Student's t- test

We formulate null and alternate hypotheses are

H0 : u1 - u2= 0 against Ha: u1 - u 2≠0

The Significance level alpha is chosen to be ∝ = 0.05

The critical region t ≥ t (0.025, 16) = ± 2.120

Degrees of freedom is calculated df = υ= n1+n2- 2= 10+8-2= 16

Here the difference between the sample means is x`1- x`2

=2902.8-3108.125

= -205.325

The pooled estimate for the common variance σ² is

Sp² = (n1-1) s1² + (n2-1) s2²/ n1+n2-2

= 9( 277.265)² + 7(3108.125)²/ 16

Sp² = 691,882.922+67,623,087.109/16

Sp= 2066.321

The test statistic is

t = (x`1- x` ) /. Sp √1/n1 + 1/n2

t= -205.325/ 2066.321 √1/10+ 1/8

t= -205.325/980.142

t= -0.2095= -0.21

Conclusion

The calculated value of t= -0.21 does not lie in the critical region t= ± 2.120 Therefore we accept our null hypothesis that there's no difference between strength (psi) of liner specimens when a certain fusion process was used and when this process was not used at the 0.05 significance level

5 0

3 years ago

PLEASE HELP

Alexeev081 [22]

Answer:

The last one is 4(6)(18).

Please give Brainliest

4 0

3 years ago

Read 2 more answers

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