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3 years ago

11

The graph represents part of Brooke's trip to work. During what time period is her speed decreasing?

2 answers:

skelet666 [1.2K]3 years ago

7 0

Answer:

B) between 20 and 25 minutes

Step-by-step explanation:

First, note that the line from 0 minutes to 5 minutes increases 40 mph. (+40)

Next, note that the line from 5 minutes to 10 minutes increases 10 mph. (+10)

Next, note that the line continues to stay the same from 10 minutes to 20 minutes. (+0)

Now, we get to the last part, which shows a drop from 50 mph to 40 mph (50 - 40 = 10; -10 mph drop)

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We can note that her speed starts at 40 mph, but <em>drops to 10 mph </em>at 5 - 10 minutes. However, note that there is no answer choice for it, and so the next logical choice is the <em>drop of speed from 20 - 25</em>, which is a 10 mph drop.

Dafna11 [192]3 years ago

3 0

B. Between 20 and 25 minuets

Brooke's speed is decreasing between 20 and 25 minutes. According to the graph Brooke's speed is decreasing during that time period.

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2 years ago

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3 years ago

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2 years ago

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What is the least common denominator for these two rational expressions n^4/n^2+2n+1,7/n^2-8n-9

sergeinik [125]

Answer:

Step-by-step explanation:

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2 years ago

What is the perimeter of triangle ABC ?

blagie [28]

Answer:

$9+\sqrt{41}$

Step-by-step explanation:

From C-A, it goes from y=1 to y=5, so that 4 units

From A-B, it goes from x = -1 to x = 4, that is 5 units

Now, to find distance from B to C, we need to use the distance formula:

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where the variables are the respective points of B and C,

B (4,5) & C(-1,1)

So x_1 =4, y_1=5, x_2=-1, y_2=1

Plugging into the formula we get:

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\D=\sqrt{(1-5)^2+(-1-4)^2}\\D=\sqrt{16+25}\\ D=\sqrt{41}$

Summing it all (perimeter is sum of 3 sides):

Distance = $4+5+\sqrt{41}\\ =9+\sqrt{41}$

3rd answer choice is right.

3 0

3 years ago