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Alika [10]
2 years ago
8

Write an equation is point slope form of a line given that the slope is 2/3 and that the line

Mathematics
1 answer:
a_sh-v [17]2 years ago
6 0
Y-y1=m(x-x1)
y-(-2)=2/3(x-5)
y+2=2/3x-10/3
y=2/3x-10/3-2
=> y=2/3x-16/3
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Anusha is walking on a hiking trail at a rate of Three-fourths miles in One-fourth hour. At this rate, how far will Anusha walk
grandymaker [24]

Answer:

Step-by-step explanation:

3/4 + 3/4 + 3/4 + 3/4 = 12/4 = 3. She will walk 3 miles in an hour.

6 0
3 years ago
What is the shorter length 60in or 8 feet
Kryger [21]
1 foot = 12 inches

12 * 5 = 60 and 8 * 12 = 96

Therefore, 60 inches is a shorter length than 8 feet.
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Solve for x in the equation x^2-4x-9=29
Kipish [7]
X = 2 + √42,2- √42
decimal form: 8.480741, -4.480741
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3 years ago
What is the slope of the line that passes through the points (9, 4)(9,4) and (3, 9)(3,9)? Write your answer in simplest form.
nalin [4]

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Step-by-step explanation:

Hope this helps <3

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2 years ago
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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
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