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____ [38]
3 years ago
14

Sam and Bobby want to know who cycled faster. The table shows the total miles Sam traveled over time. The graph shows the same r

elationship for Bobby. Who cycled faster.
Find the unit rate for Sam
Find the unit rate for Bobby
The unit rate is
Who cycled faster

Mathematics
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

<h2>Sam cycled fast, at a rate of 10 miles per hour.</h2>

Step-by-step explanation:

To solve this problem we have to find the slope of each case. The definition of a slope is:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\

Where (x_{1};y_{1}) is the first point, and (x_{2};y_{2}) is the second point.

Let's find each slope.

<h3>Sam.</h3>

Let's use the points (2;20) and (5;50)

Applying the definition of the slope, we have:

m_{sam} =\frac{50-20}{5-2}=\frac{30 \ miles}{3 \ hour}=10 \ miles/hour

This relation means that Sam cycled 10 miles per hour.

<h3>Bobby.</h3>

Let's use the points (2;18) and (6;54)

m=\frac{54-18}{6-2}=\frac{36 \ miles}{4 \ hour}=9 \ miles/hour

Bobby cycled 9 miles per hour.

Therefore, according to these ratios, Sam cycled fast, at a rate of 10 miles per hour.

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Step-by-step explanation:

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8х + 5у = -7<br> — 7х + бу = — 4
Levart [38]

Answer: x = -0.2651 and y = -0.9759

Step-by-step explanation:

Solution by substitution method

8x+5y= -7

-7x+6y= -4 = 7x-6y=4

Suppose,

8x+5y=-7→(1)

7x-6y=4→(2)

Taking equation (1), we have

8x+5y= -7

⇒8x= -5y -7

⇒x= -5y-7 / 8 →(3)

Putting x= -5y-7 / 8

in equation (2), we get 7x-6y=4

7(-5y-7/8)-6y=4

⇒-35y-49-48y=32

⇒-83y-49=32

⇒-83y=32+49

⇒-83y=81

⇒y= 81 / -83

⇒y= -0.9759→(4)

Now, Putting y= -0.9759 in equation (3), we get

x= -5y-7

x= -5(-0.9759)-7/8

⇒x= 4.8795-7/8

⇒x= -2.1205/8

⇒x= -0.2651

∴x= -0.2651 and y = -0.9759

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Farmer has 30 animals with 66 legs how many cows and chichkens
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Answer:

27 Chickens and 3 Cows.

Step-by-step explanation:

27*2=54 (chickens), 3*4=12 (cows), 54+12=66.

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Dahasolnce [82]

Answer: A x=1/4y^2

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Because to make sure on the graph is counterclockwise for the vertex for the starting point.

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