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omeli [17]
3 years ago
10

[cached] 0.020980834960938 ms

Mathematics
1 answer:
evablogger [386]3 years ago
8 0

Answer:

Just hoping this helps.

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olga_2 [115]
X + 5 < 533         Subtract 5 from each side
    - 5     -5
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x < 528
4 0
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

5 0
2 years ago
Y=<br> Pls help solve for
miss Akunina [59]

Answer:

y=39\degree

Step-by-step explanation:

The given triangle is a right angle triangle .

The known sides of the triangle that is opposite to angle y is 8 units while the side adjacent is 10 units.

The value of y can be calculated using the tangent ratio.


\tan(y)=\frac{8}{10}

We solve for y to get,

y=arctan(\frac{4}{5})

We evaluate to obtain,

y=38.66


y=39\degree to the nearest degree


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3 years ago
ANSWER QUESTION PLEASE!!
Fynjy0 [20]

Answer:

W = 2t + 8

Step-by-step explanation:

weight from birth plus number of months times 2 equals your total weight.

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3 years ago
I need to find measure of angle "u" m
garik1379 [7]

Answer:

30°

Step-by-step explanation:

TW UV

If an ARC subtends an angle in the circle (on the opposite side of the circumference), the measure of that angle will be HALF of the Arc. In other words, the Arc's measure will be DOUBLE of that angle.

The measure of the angle would be SAME if the angle subtended in the center of the circle, thought.

Now, looking at the image, we can see that:

Arc TW subtends Angle V and Angle U

*on the opposite side of the circumference*

So,

Angle V (measure given to be 30) is HALF of Arc TW, so

Arc TW = 2 * 30 = 60

So, Angle U is half of Arc TW

60/2 = 30

Hence,

<u>∠U = 30°</u>

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