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kupik [55]
3 years ago
7

SOLVE THIS PROBLEM, PLEASE! :(

Mathematics
1 answer:
pashok25 [27]3 years ago
8 0

Answer:

1st Blank: 5

2nd Blank: 3

3rd Blank: -8

4th Blank: -8

5th Blank: 12

Step-by-step explanation:

15x+35y=-100

-15x+9y=-252

___________

44y=-352

___   ___

44       44

y=-8

3x+7(-8)=-20

3x-56=-20

   +56  +56

__________

3x=36

__  __

3     3

x=12

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pshichka [43]

Answer:

We can see that the number in the middle is 36. Hence the median of the data set is 36 the median of a data set is the number placed at the middle of a data after rearrangement either in ascending order or descending order.

Given the dataset 10, 15, 27, 33, 33, 36, 42, 47, 45, 56, 78

Step-by-step explanation:

36

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Answer:

It's c) y = 15x +200

Step-by-step explanation:

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In the options given, there is just one equation that have 15 multipling x, so that must be the answer

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3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

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2 years ago
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Answer:

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6/2 = 3 so the radius = 3m

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Remember we do not apply pi because the answer should be in terms of pi.

Now we subtract the two

Area of white portion = 49 - 9π

Hence, the answer is A

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3 years ago
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