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adoni [48]
2 years ago
5

What is limit of (x cubed 3 x squared 4 x minus 12) as x approaches negative 3? –78 –24 30 54

Mathematics
1 answer:
Sergio039 [100]2 years ago
7 0

The limit of the expression as x approaches -3 is -24

<h3>How to determine the limit of the expression?</h3>

The expression is given as:

x^3 + 3x^2 + 4x - 12

As x approaches -3.

The limit expression becomes

\lim_{x \to -3} x^3 + 3x^2 + 4x - 12

Substitute -3 for x in the expression

\lim_{x \to -3} (-3)^3 + 3(-3)^2 - 4*3 - 12

Evaluate the expression

\lim_{x \to -3} -24

Hence, the limit of the expression as x approaches -3 is -24

Read more about limit expressions at:

brainly.com/question/16176002

#SPJ4

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There were 60 runners to start the race. In the first half of the race,1/3 of them dropped out. In the second half of the race,1
Verizon [17]
In the question, it is already given that the total number of runners in the race is 60 and out of them 1/3 dropped out in the first half. In the second half 1/4 of the remaining runners dropped out.
Now
Total number of runners in the race = 60
Number of runners that dropped out in the first half = 1/3 * 60
                                                                                   = 20
Number of runners remaining = 60 - 20
                                                 = 40
Number of runners dropping out in the second half = 40 * 1/4
                                                                                 = 10
Then the number of runners that finished the race = 40 - 10
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5 0
2 years ago
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At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
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