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IceJOKER [234]
2 years ago
14

dle" class="latex-formula"> + \frac{3}{10} + \frac{3}{5} = __
(it would help if you showed how you did it)
Mathematics
2 answers:
Kisachek [45]2 years ago
8 0
Hey, I hope this helps! I tried to write down some stuff for you.

aleksklad [387]2 years ago
3 0
So what you do,is you add them together to get 13/10 but then you can also simplify it but you can keep the answer to 13/10
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Nutka1998 [239]

Answer:

but i think smar 5555

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3 years ago
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
nata0808 [166]

<em>Answer</em>

<h2><em>6</em><em>9</em><em>.</em><em>4</em><em>9</em><em>8</em><em> </em><em>cm^</em><em>2</em></h2>

<em>diameter=</em><em>1</em><em>8</em><em> </em><em>cm</em>

<em>radius=</em><em>1</em><em>8</em><em>/</em><em>2</em><em>=</em><em>9</em><em>c</em><em>m</em>

<em>Area </em><em>of </em><em>circle=</em><em> </em><em>pi </em><em>r^</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>3</em><em>.</em><em>1</em><em>4</em><em>2</em><em>*</em><em>(</em><em>9</em><em>)</em><em>^</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>2</em><em>5</em><em>4</em><em>.</em><em>5</em><em>0</em><em>2</em><em> </em><em>cm^</em><em>2</em>

<em>Area </em><em>of </em><em>square=</em><em> </em><em>(</em><em>l)</em><em>^</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>(</em><em>1</em><em>8</em><em>)</em><em>^</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>3</em><em>2</em><em>4</em><em> </em><em>cm^</em><em>2</em>

<em>Total </em><em>shaded </em><em>area=</em><em> </em><em>3</em><em>2</em><em>4</em><em>-</em><em>2</em><em>5</em><em>4</em><em>.</em><em>5</em><em>0</em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>6</em><em>9</em><em>.</em><em>4</em><em>9</em><em>8</em><em> </em><em>cm^</em><em>2</em>

<em>Hope </em><em>it</em><em> helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

7 0
3 years ago
Identify the inequalities A, B, and C for which the given ordered pair is a solution: (0,0)<br>​
professor190 [17]

Answer:

Options (A) and (C)

Step-by-step explanation:

A). The given inequality is,

    x + y ≤ 2

    Solution area of the inequality is the shaded area below the line.

    Since (0, 0) lies in the shaded region, ordered pair will be a solution of this inequality.

B). For the inequality y ≤ \frac{3}{2}x-1

   Since, (0, 0) doesn't lie in the shaded region of this inequality, ordered pair will not be a solution of the inequality.

C). y > -\frac{1}{3}x-2

    In this graph (0, 0) lies in the shaded region of the inequality therefore, it will be a solution of the given inequality.

Therefore, Options (A) and (C) are the correct options.

5 0
3 years ago
A random sample of 50 bottles is selected from the production line of a large manufacturing company. The mean weight of the cont
andrew11 [14]

Answer:

The population is all bottles in the production line and the sample is the 50 bottles that were selected.

Step-by-step explanation:

Edge 2020

4 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
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