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3 years ago

Find the precentage 87% of 55

2 answers:

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

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

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

=> $25y^2 = 9y^2 + 9 * 200^2$

=> $16y^2 = 360000$

=> $y = 150$

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 following graphs: 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.

There are 14 vertices for 105 edges
There are 200 vertices for 19900 edges

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

So, we represent the edges as:

$E \to Edges$

(a) The value of N for 105 edges

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:

$N = 15$

(b) The value of N for 19900 edges

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:

$N = 200$

Hence, there are 200 vertices for 19900 edges

Read more about vertices and edges at:

brainly.com/question/22118318

7 0

3 years ago