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Ivanshal [37]
1 year ago
6

If another line is perpendicular to the line shown on the graph, what must be its slope

Mathematics
1 answer:
Shtirlitz [24]1 year ago
4 0

It's B

Slope is negative, the other must be positive due to m1 x m2 = - 1

m1 = -2/5

The diagram shows a gradient of -2/5. So, the

reciprocal of this valve to make - 1 is 5/2 (which is the - 2/5 flipped upside down, but positive)

-2/5 x 5/2 = - 10/10 (or - 1)

Thus, the perpendicular line to this has a gradient of 5/2.

Hope this helps!

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Median AM and CN of △ABC intersect at point O. What part of area of △ABC is the area of △AON?
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Answer:

The area of △AON is one-sixth part of △ABC.

Step-by-step explanation:

Let the total area of △ABC be x.

Median divides the area of a triangle in two equal parts.

Since AM are CN are medians, therefore the area of △ACN, △BCN, △ABM and △ATM are equal, i.e., \frac{x}{2}.

The intersection point of medians is called centroid of the triangle. A centroid divides the median in 2:1.

Since CN is median and O is the centroid of the triangle, therefore CO:ON is 2:1.

Draw a perpendicular on CN from A.

\frac{\text{Area of }\triangle ANO}{\text{Area of }\triangle ACN}=\frac{\frac{1}{2}\times ON\times h}{\frac{1}{2}\times CN\times h}=\frac{1}{3}

Therefore the area of △AON is one-third of △ACN.

\text{Area of }\triangle ANO=\frac{1}{3}\times\text{Area of }\triangle ACN

\text{Area of }\triangle ANO=\frac{1}{3}\times\frac{x}{2}

\text{Area of }\triangle ANO=\frac{x}{6}

The area of ANO is \frac{x}{6}. Therefore the area of △AON is one-sixth part of △ABC.

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Kevin can travel 22 1/2 miles in 1/4 hours . What is his average speed in miles per hour
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A weight-loss clinic advertised that last month its patients, on average,8lb lost . Assuming that average refers to the mean, wh
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Answer:

Given:

A weight-loss clinic advertised that last month its patients, on average, lost 8 lbs

Average refers to mean

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Answer: Mean or average is a measure of central tendency or in other terms refers to the central value of the dataset. If the value of mean/average is 8lb, it signifies that there are some patients who have lost weights more than 8 lbs. So, this statement is False

b. Last month at least one of their patients lost exactly 8 lbs.

Answer: It is a possibility that all the patients have lost their weights greater than or less than 8 lbs. Since the mean is computed by summing up all the values (weight loss in this case) by the total number of observations, it is possible that there can be no patient who has lost exactly 8lbs. So, this statement is False.

c. Two months ago some of their patients lost at least 8 lbs.

Answer: It is clearly mentioned that last month its patients, on average, lost 8 lbs. No information is given about the weight loss of patients before the last month. So, there is not sufficient data to check whether this statement is correct or not.

d. Last month at least one of their patients lost less than 3 lbs.

Answer: It is a possibility that all the patients have lost their weights greater than 3 lbs. Since the mean is computed by summing up all the values (weight loss in this case) by the total number of observations, it is possible that there can be no patient who has lost less than 3lbs. So, this statement is False.

e. Last month, the number of their clients who lost less than 8 lbs was equal to the number of their clients who lost more than 8 lbs.

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Answer e.Last month, the number of their clients who lost less than 8 lbs was equal to the number of their clients who lost more than 8 lbs is True

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Help with #10 plz...
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A. is 21.98 cm
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