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almond37 [142]
3 years ago
9

Can someone​ please help me I really need it

Mathematics
2 answers:
Andreas93 [3]3 years ago
8 0
Check the picture below

notice, both triangles are 45-45-90 triangles, so if one has a base of 26, so does the other

Lesechka [4]3 years ago
4 0
Download the app Slater! You put in the title of any text book and it'll pop up.
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For functions f(x) = x + 2 and g(x) = 8x2 - 2x + 4,
adoni [48]

Answer:

Sorry im not ood at math so i cant help

Step-by-step explanation:

4 0
3 years ago
Is (a + b) and (a - b)? perfect square trinomials?<br> O True<br> O False
lisabon 2012 [21]

Answer:

False

Step-by-step explanation:

Pos and negs cannot be perfect square trinomials

8 0
3 years ago
Read 2 more answers
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
Tim​ Urban, owner/manager of​ Urban's Motor Court in Key​ West, is considering outsourcing the daily room cleanup for his motel
pantera1 [17]

Answer:

broke

Step-by-step explanation:

----$$$$$$$

7 0
3 years ago
Use the function below to find F(3)
garik1379 [7]

Answer:

Theanswer is 4/27.

Step-by-step explanation:

given that, F(x) = 4×(1/3)^x

now , F(3)= 4×(1/3)^3 ( putting value of x)

or, F(3) = 4×(1/27)

therefore, F(3)= 4/27... ans

<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

8 0
3 years ago
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