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ArbitrLikvidat [17]
2 years ago
9

I cant figure out what is 4×(3x+5

Mathematics
2 answers:
adoni [48]2 years ago
6 0

Answer:

12x+20

Step-by-step explanation:

Just expand the brackets

damaskus [11]2 years ago
6 0

you have to do 12x+20.

hope this works

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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

5 0
2 years ago
Simplify the algebraic expression (8x + 16y)/2 + 4(x - y)
Karo-lina-s [1.5K]

It's 8x + 16y/2 + 4x - 4y :))

5 0
2 years ago
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Solve this equation 7x+3=5(x-3)+2x
zaharov [31]

Answer: nonsense

Step-by-step explanation:

7x+3=5(x-3)+2x

7x+3=5x-15+2x

7x+3-5x+15-2x=0

18=0

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8 0
2 years ago
Does a polygon usually have more sides or <br> more angles
xxTIMURxx [149]

I've seen this question on Brainly before, and I always shake my head.

Please think about this for a few seconds.  Maybe even make some
scribbles on a piece of paper.

-- A triangle has 3 sides and 3 angles.

-- A square, rectangle, rhombus or parallelogram has 4 sides and 4 angles.

-- Draw anything with 5 sides.  It doesn't have to be pretty, and they don't
all have to be the same length or anything special.  Just draw any shape
with 5 sides.  Count the angles, and you'll find that there are five of them.

By now you should be starting to get the creepy hunch that maybe a
polygon always has the SAME number of sides and angles.  I hope so. 
That's the correct creepy hunch. 

You can get all kinds of hunches, and even work most of them out,
just by using your thinker for a while.


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3 years ago
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49 and 7\12 as a mixed number
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The number can be rewritten as;
49+7/12
49\frac{7}{12}
8 0
3 years ago
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