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nadezda [96]
3 years ago
5

0.5x+7/3 = 21/4 cross multiply

Mathematics
1 answer:
TiliK225 [7]3 years ago
3 0

Solving the expression  \frac{0.5x+7}{3}=\frac{21}{4} we get: x=\frac{35}{2}\,\,or\,\,x=17.5

Step-by-step explanation:

Solving the expression: \frac{0.5x+7}{3}=\frac{21}{4}

Solving by cross multiplying:

\frac{0.5x+7}{3}=\frac{21}{4}\\(0.5x+7)*4=21*3\\Simplifying:\\2x+28=63\\Adding\,\,-28\,\,on\,\,both\,\,sides:\\2x+28-28=63-28\\2x=35\\Divide\,\,both\,\,sides\,\,by\,\,2\\\frac{2x}{2}=\frac{35}{2}\\x=\frac{35}{2}\,\,or\,\,x=17.5

So, solving the expression  \frac{0.5x+7}{3}=\frac{21}{4} we get: x=\frac{35}{2}\,\,or\,\,x=17.5

Keywords: Solving Fractions

Learn more about Solving fractions at:

  • brainly.com/question/2456302
  • brainly.com/question/1648978
  • brainly.com/question/13174289
  • brainly.com/question/1677114

#learnwithBrainly

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Step-by-step explanation:

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2 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)

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Citrus2011 [14]

Answer:

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Step-by-step explanation:

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2 years ago
Let L be the line parametrized by x = 2 + 2t, y = 3t, z = −1 − t. (a) Find a linear equation for the plane that is perpendicular
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Answer:

Step-by-step explanation:

Given that L is a line parametrized by

x = 2 + 2t, y = 3t, z = −1 − t

The plane perpendicular to the line will have normal as this line and hence direction ratios of normal would be coefficient of t in x,y,z

i.e. (2,3,-1)

So equation of the plane would be of the form

2x+3y-z =K

Use the fact that the plane passes through (2,0,-1) and hence this point will satisfy this equation.

2(2)+3(0)-(-1) =K\\K =5

So equation is

2x+3y-z =5

b) Substitute general point of L in the plane to find the intersecting point

2(2+2t)+3t-(-1-t) =5\\4+8t+1=5\\8t=0\\t=0\\(x,y,z) = (2,0,-1)

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3 years ago
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