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Vitek1552 [10]
3 years ago
9

What is the simplified value of the expression below?

Mathematics
2 answers:
bagirrra123 [75]3 years ago
7 0
1/2
Is the answer
To the problem
ankoles [38]3 years ago
7 0

Answer:

option C : 1/2

Step-by-step explanation:

\frac{1}{3} \div \frac{2}{3} \\\\\frac{\frac{1}{3}}{\frac{2}{3}}\\\\\frac{1}{3} \times \frac{3}{2} \\\\\frac{1}{2}

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Complete the sentence. 22/7 divided by 2 is ? The value of (22/7 minus 2/3)
MakcuM [25]

Answer:

11/7

Step-by-step explanation:

22/7 divided by 2 = 22/7 multiplyed by 1/2 = 22/14 = 11/7

I don't really know what the second sentence means, so plz tell me, but here 22/7 minus 2/3 is 52/21

4 0
3 years ago
(Help me please) what can you tell about the mean of each distribution
vfiekz [6]

Answer:

The distribution is considered the mean in the problem

Step-by-step explanation:

$124.84 - $19.75 = <u>$105.09</u>

<u>If you can consider what the mean is in math: How to Find the Mean. The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.</u>

<h2><u>Have a nice day and if you could mark me as brainllest I would be happy</u></h2>
8 0
3 years ago
Read 2 more answers
30 POINTS!! WILL VOTE BRAINIEST
Alex787 [66]
Part A:

Consider from x = -5 to x = -4, they are 1 unit apart and the difference of their outputs is given by:

-3 - (-11) = -3 + 11 = 8.

Thus, the value of the output increases by 8 units for each one unit increase in the input.



Part B:

Consider from x = -3 to x = -1, they are 2 units apart and the difference of their outputs is given by:

21 - 5 = 16.

Thus, the value of the output increases by 16 units for each two units increase in the input.



Part C:

Consider from x = 0 to x = 3, they are 3 units apart and the difference of their outputs is given by:

53 - 29 = 24.

Thus, the value of the output increases by 24 units for each three units increase in the input.



Part D:

It can be noticed that the ratio difference in the outputs to the input intervals are equal for all the given input intervals.

i.e 8 / 1 = 16 / 2 = 24 / 3.
7 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
Please help please please
Murrr4er [49]
2 pages per hour.

the fractions can be turned to decimals easily, 4/5 is 0.8 and 2/5 is 0.4 . divide those decimals together and you get 2.
4 0
3 years ago
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