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Snowcat [4.5K]
3 years ago
7

A bug is moving back and forth on a straight path. The velocity of the bug is given by vt=t2-3t. Find the average acceleration o

f the bug on the interval 1, 4.
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

2 unit/time²

Step-by-step explanation:

Given the equation:

v(t) =t^2-3t

At interval ; 1, 4

V(1) = 1^2 - 3(1)

V(1) = 1 - 3

V(1) = - 2

At t = 4

V(4) = 4^2 - 3(4)

V(4) = 16 - 12

V(4) = 4

Average acceleration : (final - Initial Velocity) / change in time

Average acceleration = (4 - (-2)) ÷ (4 - 1)

Average acceleration = (4 + 2) / 3

Average acceleration = 6 /3

Average acceleration = 2

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85 / 5 

(50 + 35) / 5 =
(50/5) + (35/5) =
10 + 7 = 17
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3 years ago
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Determine whether Point A lies on the circle whose center is Point C and whichcontains the Point P. Justify your answer mathemat
Arada [10]

The Solution:

Given:

Center = (0,0)

Point A = (-5,2) being a point on the circle.

We are required to check if point P = (2,-5) is on the circle.

Solving the given problem graphically, we have:

From the above graph, it is clear that point P(2,-5) is a point on the circle.

8 0
11 months ago
36 creates of apples. each creat has 25 kgs of apples. they sell 486,235 games of apples. how many grams are left?
valentinak56 [21]

The grams of apples left is 413765 grams

<h3>How to determine the value</h3>

From the information given, we have that:

  • There are 36 crates of apples
  • Each contains 25 kilograms of apples
  • 486,235 grams of apples were sold

If we have 36 crates, let's determine the the total grams of apples

1 crate = 25 kilograms = 25000 grams

36 crates = x

x = 25000 × 36

x = 900000 grams

The total grams of apple is  900000 grams

The grams left = Total - sold =  900000 - 486,235 = 413765 grams

Thus, the grams of apples left is 413765 grams

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8 0
2 years ago
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Blababa [14]
Oh yeah sweetie no I’m not here for church I just got a text message I don’t think I can
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3 years ago
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An object at the top of a building with height 110 feet is thrown upward with an initial speed of 23 ft/s. Find its position z(t
alina1380 [7]

Answer:

Step-by-step explanation:

Given

Height of building is h=110\ ft

Initial upward velocity u=23\ ft/s

height of object is given by

z(t)=ut+\frac{1}{2}gt^2

but building is already at a height of 110 ft so z(t) must be given by

Z(t)=ut+\frac{1}{2}(-g)t^2+110

Z(t)=23\times t-\frac{1}{2}\times 9.8\times t^2

Z(t)=-16t^2+23t+110

Time when it hits is given by when Z(t)=0[/tex]

-16t^2+23t+110=0

t=\frac{-23\pm \sqrt{(23)^2-4\times (-16)\times 110}}{2\times (-16)}

t=\frac{-23\pm 87}{-32}

t=3.43\ s

6 0
3 years ago
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