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3 years ago

Please help me with algebra! Don't answer if you don't know how to do it.

Mathematics

1 answer:

astra-53 [7]3 years ago

3 0

This is all a bit complicated so try and stick with me on this one!

This one is the first problem in the first picture! a + 2 / a^2 + a - 1 / a + 1/ -a^2 - 2a - 1

= a^4 + a^3 - a^2 - a / -a^5 - 3a^4 - 3a^3 - a^2

= -a^3 - a^2 + a + 1 / a^4 + 3a^3 + 3a^2 + a

= -a^3 - a^2 + a + 1/ a^4 + 3a^3 + 3a^2 + a

= (-a - 1) (a + 1) (a - 1) / a(a+ 1) (a + 1) (a + 1)

= -a + 1 / a^2 + a

The second problem in the first picture! 3x / y + 3x - y^2 / 3xy - 9x^2 + y^2 + 9x^2 / y^2 - 9x^2

= -81x^4y + 18x^2y^3 - y^5 / 243x^5 - 54x^3y^2 + 3xy^4

This one is for the last picture! 4y^2 + 4y + 1 / 4y - 8y^2 - 4y^2 + 1 / 4y + y

= 16y^3 + 24y^2 / -32y^3 + 16u^2

= 16y + 24 / -32y + 16

= 2y + 3 / -4y + 2

I hope this was helpful!!!

Please remember to rate/thank/mark as brainliest if you see fit! (:

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Answer:

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Step-by-step explanation:

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1 year ago

The rectangular bar is connected to the support bracket with a 10-mm-diameter pin. The bar width is w = 70 mm and the bar thickn

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Answer:

P_max = 9.032 KN

Step-by-step explanation:

Given:

- Bar width and each side of bracket w = 70 mm

- Bar thickness and each side of bracket t = 20 mm

- Pin diameter d = 10 mm

- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa

- Average allowable shear stress of pin S = 115 MPa

Find:

The maximum force P that the structure can support.

Solution:

- Bearing Stress in bar:

T = P / A

P = T*A

P = (120) * (0.07*0.02)

P = 168 KN

- Shear stress in pin:

S = P / A

P = S*A

P = (115)*pi*(0.01)^2 / 4

P = 9.032 KN

- Bearing Stress in each bracket:

T = P / 2*A

P = T*A*2

P = 2*(120) * (0.07*0.02)

P = 336 KN

- The maximum force P that this structure can support:

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P_max = 9.032 KN

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3 years ago

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Step-by-step explanation:

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2 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

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Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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