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Verdich [7]
3 years ago
6

Find the precentage 87% of 55

Mathematics
2 answers:
kotegsom [21]3 years ago
4 0
The answer is 47.85 because you're just multiplying the percentage to the given number, but first, convert the percentage into a decimal.
nirvana33 [79]3 years ago
3 0
87% equals to .85

.87 • 55= 47.85
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Como resuelvo este ejercicio//-7-x+5=-3+5
Snezhnost [94]

Answer:

x=-4

Answer here

5 0
3 years ago
Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin loc
pshichka [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The optimal point(i.e the position of  P) that will allow Sally to reach the cabin as quickly as possible is  d =  50 \ ft from the cabin.

Step-by-step explanation:

From the question we are told that

   The speed from the point Sally is standing to the point P is  u =  3 \  ft /s

   The speed from the point P to the cabin is v  =  5 \  ft /s

   

Let denote the distance from the bottom left corner to the  P  be  y  ft

 Generally according to Pythagoras theorem the distance from Sally's first position to point P is mathematically represented as

          s = \sqrt{y^2 + 200^2}

Generally the distance from that point P to the cabin is mathematically represented as

       d =  200 -y

Generally the time it takes Sally to cover that distance s is mathematically represented as

       t_1 =  \frac{s}{u}

=>     t_1 =  \frac{ \sqrt{y^2 + 200^2}}{3}

Generally the time it takes Sally to cover that distance d is mathematically represented as

       t_2 =  \frac{d}{v}

=>     t_2 =  \frac{ 200 - y }{5}

So the total time taken is  

        t= \frac{ \sqrt{y^2 + 200^2}}{3}  +  \frac{ 200 - y }{5}

Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0

So  

     \frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} }  -  \frac{1}{5}  =0      

=> \frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}

=> 25y^2 = 9y^2 + 9 * 200^2        

=> 16y^2 = 360000          

=> y = 150

So the distance from point P to the cabin is  

           d =  200 -150

=>       d =  50 \ ft

So  the optimal point(i.e the position of  P) that will allow Sally to reach the cabin as quickly as possible is  d =  50 \ ft from the cabin

       

8 0
2 years ago
Identify the constant of proportionality in the<br> following graphs:<br><br> Plz help
Sedbober [7]

Answer:

its proportionality

Step-by-step explanation:

8 0
3 years ago
Joseph ran LaTeX: 2\frac{3}{4}2 3 4 miles in LaTeX: \frac{2}{5}2 5 of an hour. How fast did Lin run in 1 hour? (try to leave you
Olegator [25]

Answer:

Nathaniel ran fastest

Step-by-step explanation:

For Joseph

Joseph ran 2 3/4 miles 2 /5 of an hour.

Speed = Distance/Time

Speed = 2 3/4 miles ÷ 2/5 hour

= 11/4 ÷ 2/5

= 11/4 × 5/2

= 55/8 miles/hour or 6.875 miles/hour

How fast did he run in 1 hour? (try to leave your answer in a fraction)

Hence, this is calculated as:

2/5 hour = 55/8 miles per hour

1 hour = x

Cross Multiply

x = 55/8 ÷ 2/5

x = 55/8 × 5/2

x = 275/16 miles per hour

x = 17 3/16 miles per hour

Nathaniel ran LaTeX: 8 2/3 miles in 4 /3 of an hour. How fast did Nathaniel run in 1 hour? (try to leave your answer in a fraction)

Speed = Distance/Time

Speed = 8 2/3 miles ÷ 4/3hour

= 26/3 ÷ 4/3

= 26/3 × 3/4

= 26/4 miles/hour

How fast did he run in 1 hour? (try to leave your answer in a fraction)

Hence, this is calculated as:

4/3 hour = 26/4miles per hour

1 hour = x

Cross Multiply

x = 26/4 ÷ 4/3

x = 26/4 × 3/4

x = 78/16 miles per hour

x = 4 14/16 miles per hour

= 4 7/8 miles per hour

Who ran the fastest, Nathaniel or Joseph?

Hence, from the calculation above, Nathaniel ran fastest

5 0
3 years ago
A) Find the value of N if, Kn has 105 edges.
blagie [28]

The number of edges can be calculated from the number of vertices.

  • <em>There are 14 vertices for 105 edges</em>
  • <em>There are 200 vertices for 19900 edges</em>

The variable N is used to always represent the number of vertices.

So, we represent the edges as:

E \to Edges

<u />

<u>(a) The value of N for 105 edges</u>

The relationship between N and E is:

E = \frac{N \times (N - 1)}{2}

Substitute 105 for E

105 = \frac{N \times (N - 1)}{2}

Multiply through by 2

210 = N \times (N  - 1)

210 = N^2  - N

Rewrite as:

N^2 - N - 210 = 0

Expand

N^2 +14N - 15N - 210 = 0

Factorize

N(N +14) - 15(N + 14) = 0

Factor out N + 14

(N - 15) (N + 14) = 0

Solve for N

N = 15 or N = -14

The number of vertices (N) cannot be negative. So:

N = 15

<u>(b) The value of N for 19900 edges</u>

We have:

E = \frac{N \times (N - 1)}{2}

Substitute 19900 for E

19900 = \frac{N \times (N - 1)}{2}

Multiply through by 2

39800 = N \times (N  - 1)

39800= N^2  - N

Rewrite as:

N^2 - N - 39800= 0

Expand

N^2 +199N - 200N - 39800= 0

Factorize

N(N +199) - 200(N + 199) = 0

Factor out N + 199

(N + 199) (N - 200) = 0

Solve for N

N = 200 or N = -199

The number of vertices (N) cannot be negative. So:

N = 200

<em>Hence, there are 200 vertices for 19900 edges</em>

Read more about vertices and edges at:

brainly.com/question/22118318

7 0
3 years ago
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