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Hitman42 [59]
3 years ago
11

What is the solution of the equation sqrt 2 x+4-sqrt x=2

Mathematics
1 answer:
ivann1987 [24]3 years ago
3 0
There are 2 answers to this 

x=0 
x=16
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Write the rational number 0.3 in the form ab , where a and b are integers.
grandymaker [24]

Answer: 3/10

Step-by-step explanation:

Given a decimal number 0.3

To Write the rational number 0.3 in the form ab , where a and b are integers, take it to be :

0.3 /1

Multiply the numerator and the denominator by 10

0.3/1 × 10/10 = 3/10

Therefore, a = 3 and b = 10

the rational number 0.3 in the form ab , where a and b are integers is

3/10

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3 years ago
Help I need help with this one
andrey2020 [161]
THE CORRECT ANSWER IS D 1/6
8 0
3 years ago
A stone dropped from a bridge strikes the water 2.2 s later. How high is the bridge above thewater
vivado [14]
Let the only force acting on the ball is the gravity (free-failing)  and the ball was initially stationary
vo = 0

s = vot + 1/2 gt^2

s = 1/2 x 10 x (2.2) ^2 = 24.2 m 
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3 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
One batch of cookies makes 18. How many
Viktor [21]

Answer:

6

Step-by-step explanation:

100/18=5 1/2

18*6=108

6 0
2 years ago
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