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harina [27]
3 years ago
5

My base is 10. My exponent is an odd number. If you write a multiplication sentence to show my value, it will have fewer than 4

factors of 10. My value is greater than 10. What number am I?
Mathematics
1 answer:
svet-max [94.6K]3 years ago
4 0
. The number is either 10^1 or 10^3 or 10^ 5 or . . . . 

<span>If it is more than 10, then it is not 10^1. The next possibility is 10^3which is 1000, which has factors 2, 5, 10, 20, 25, 40, 50, 80, . . . . . </span>

<span>What constitutes a "multiplication sentence" exactly ? This is not a standard mathematical term. Do you mean "reduce it to its prime factors" ? </span>

<span>1000 = 2 x 2 x 2 x 5 x 5 x 5 </span>

<span>which is a lot more than 4 factors. Or not, if you write it as 2³ x 5³. </span>

<span>But you could say the same about 100 000 = 2^5 x 5^5 </span>

<span>or about 10 000 000 = 2^7 x 5^7 </span>

<span>Take your pick. This is a very badly stated question. </span>
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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
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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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