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3 years ago

If UT is a midsegment of QRS, find SQ.

Mathematics

1 answer:

serg [7]3 years ago

8 0

If UT is a midsegment,
SR = 2·UT
7x-1 = 2(3x+1)
We can add 1-6x to solve this
x = 3

Then the length SQ is twice the length of SU.
SQ = 2(2x+2) = 2(2·3+2)
SQ = 16

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Alina [70]

$7 - 6b \leq 19 - 7 \\ \\ -6b \leq 12 \\ \\ b \geq \frac{12}{-6} \\ \\ b \geq - \frac{12}{6} \\ \\ b \geq -2 \\ \\ Answer: b \geq -2$

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3 years ago

The answer for this question cause i dont get it

ycow [4]

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3 0

3 years ago

Read 2 more answers

If the first number is increased by 30% and the second number by 80%, the sum will be 1.6 times larger than the initial sum. Fin

Bezzdna [24]

Answer:

2 : 3

Step-by-step explanation:

Let x represent the first number and let y represent the second number.

Therefore the initial sum of the numbers = x + y

Given that the first number is increased by 30%, it becomes = x + 30% of x = x + 0.3x = 1.3x.

Also, the second number is increased by 80%, it becomes = y + 80% of y = y + 0.8y = 1.8y.

The sum of this increased numbers is 1.6 times larger than the initial sum. That is:

1.3x + 1.8y = 1.6(x + y)

1.3x + 1.8y = 1.6x + 1.6y

1.8y - 1.6y = 1.6x - 1.3x

0.2y = 0.3x

The ratio of the first number to the second number (that is x / y) is:

0.3x = 0.2y

Divide through by 0.3y:

0.3x / 0.3y = 0.2y / 0.3y

x / y = 0.2 / 0.3

x / y = 2 / 3

x : y = 2 : 3

4 0

3 years ago

Can anybody help me? I thought It was D on the first problem but i'm not sure.

uysha [10]

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{small~box}{large~box}\qquad \qquad \stackrel{sides}{\cfrac{s}{s}}=\cfrac{1}{3}\qquad \qquad \stackrel{volumes}{\cfrac{s^3}{s^3}}=\cfrac{1^3}{3^3}\implies \cfrac{1}{27}$

7 0

2 years ago

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi

bazaltina [42]

Answer:

$425.6 should be budgeted for weekly repairs and maintenance.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean $400 and standard deviation $20.

This means that $\mu = 400, \sigma = 20$

How much should be budgeted for weekly repairs and maintenance to provide that the probability the budgeted amount will be exceeded in a given week is only 0.1?

This is the 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

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

$1.28 = \frac{X - 400}{20}$

$X - 400 = 20*1.28$

$X = 425.6$

$425.6 should be budgeted for weekly repairs and maintenance.

6 0

2 years ago