3 years ago

-3x+4y=-10 in y=mx+b

Mathematics

1 answer:

belka [17]3 years ago

8 0

Y = 3/4x -5/2

Hope this helps!

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What constant acceleration is required to increase the speed of a car from 30 mi/h to 52 mi/h in 5 seconds? (Round your answer t

Alenkasestr [34]

Answer:

$a = 9.84 m/s^2$

Step-by-step explanation:

Given:

Initial speed (Vi) = 30 mi/h = 67.1 m/s

Final speed (Vf) = 52 mi/h = 116.32

Time period(t) = 5 seconds

We know the formula of acceleration in terms of initial speed, final speed and time:

$a = \frac{Vf -Vi}{t}$

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$a = 9.84 m/s^2$

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dmitriy555 [2]

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The factored form of 4a3b5 − 16a5b2 + 12a2b3 is ?

Lady bird [3.3K]

4a^2b^2(ab^3-4a^3+3b)

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3 years ago

Read 2 more answers

red bike is traveling south, toward Lawrence, at constant speed of 24km/h and a blue bike is also traveling east, away from Lawr

zavuch27 [327]

Answer:

a. $R=10-24t$

b. $B=6+6t$

Step-by-step explanation:

We are given that

Speed of red bike, v=24 km/h

Speed of blue bike, v'=12km/h

Distance of red bike when t=0

=10 km north

Distance of blue bike when t=0

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a. We have to find the expression of distance of red bike from Lawrence.

$Distance=speed\times time$

Let t be the time taken by red bike

Distance of red bike from Lawrence is given by

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$R=10-24t$

Distance of the blue bike from Lawrence

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3 years ago

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Mars2501 [29]

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3 years ago