Urgentthe area of a square field is 60025m square.a man cycles along its boundary at 18 km per hour
in how much time will he return at starting point

Answers

Answer 1
Answer: Area of square field = 60025 m

Thus, side = √60025 = 245 m.

Perimeter = 245 * 4 = 980 meters

Speed = 18 kmph = 5m/s

Time = Distance / speed

= 980 / 5

= 196.

Thus, it will take 196 seconds for the man to return at the starting point.

Answer 2
Answer: Area=60025\ m^2\ \ \ and\ \ \ Area=a^2\n\n\ \ \ \Rightarrow\ \ \ a^2=60025\ m^2 \ \ \Rightarrow\ \ \ a=245\ m\n\n\ \Rightarrow\ \ \ the\ perimeter\ of\ a\ square:\n\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P=4a=4\cdot245\ m=980\ m=0.98\ km\n\n Speed=18 (km)/(h) \ \ \ and\ \ \ Speed= \frac{\big{P}}{\big{time}} \n\n18= \frac{\big{0.98}}{\big{time}}\n\n\ \ \ \Rightarrow \ \ \ time= \frac{\big{0.98}}{\big{18}}\ hr=\frac{\big{0.98}}{\big{18}}\ \cdot3660\ s=196\ s=3\ min\ 16\ s\n

Ans.\ he\ will\ return \ after\ 3\ minutes\ 16\ seconds.

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I NEED HELP FAST. DUE TOMORROW

Find the constant of proportionality and unit rate for the data in the table. Then write the equation to represent the relationship between time t and distance d Time Distance
2 hrs 90 miles
3 hrs 135 miles
5 hrs 225 miles
6 hrs 270 miles
I dont understand how to do this kind of math.Please help

Answers

You should set up an X and Y table. Make time X and Distance Y. (the table is like a cross: on the left top put x and left top put y. Then under the line list the x and y numbers given in order.) Then find the "slope" which is y/x  (rise over run) then you get the increase rate of 45 mph. The relation between them would be D=45t 

PLEASE PLEASE HELP IVE BEEN DOING THIS FOR AN HOUR​

Answers

Answer:

Use: sin

h= 78.8

Step-by-step explanation:

1 7/9 divided by 4/5 =

Answers

The quotient of dividing the given fractions is 20/9.

Given that, 1 7/9=16/9 and 4/5.

From the given fractions we need to divide 16/9  by 4/5.

How to divide the fractions?

A fraction represents a part of a whole or, more generally, any number of equal parts.

The first step to dividingfractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. Finally, simplify the fractions if needed.

Now, 16/9÷4/5

Change ÷ to × and reciprocate 4/5 to 5/4.

Then, 16/9÷4/5=16/9×5/4=20/9

Therefore, the quotient of the given fraction is 20/9.

To learn more about the division of fractions visit:

brainly.com/question/17205173.

#SPJ2

Answer:

20/9

Step-by-step explanation:

16/9 * 5/4

4/9 * 5

20/9

What is the solution to 2x+5>3x+1

Answers

   2x + 5 > 3x + 1
 - 2x        - 2x     
           5 > x + 1
         - 1        - 1
           4 > x
           x < 4

The solution is 4>x Because you subtract 1 from both sides and you subtract 2x from both sides and you get x<4 or 4>x

What is the highest common factor of 12 and 49?

Answers

12|2\n.\ 6|2\n.\ 3|3\n.\ 1|\n==========\n49|7\n.\ 7|7\n.\ 1|\n==========\n\nHCF(12;\ 49)=1
Factors of 12 . . . . . 1, 2, 3, 4, 6, 12

Factors of 49 . . . . . 1, 7, 49

Common factors . . . 1

Highest common factor . . . 1

Simplify the expression

Answers

Answer:

The simplified form of given expression\frac{15xy}{5x^{(1)/(2)}y^2} is \frac{3x^{(1)/(2)}}{y}

Step-by-step explanation:

Given: Expression \frac{15xy}{5x^{(1)/(2)}y^2}

We have to write the given expression in simplified form,

Consider the given expression \frac{15xy}{5x^{(1)/(2)}y^2}

Divide the numbers (15)/(5)=3

we get,

=\frac{3xy}{y^2x^{(1)/(2)}}

Apply exponent rule , (x^a)/(x^b)\:=\:x^(a-b)

\frac{x}{x^{(1)/(2)}}=x^{1-(1)/(2)}=x^{(1)/(2)}

we get,

=\frac{3yx^{(1)/(2)}}{y^2}

Cancel y term, we have,

=\frac{3x^{(1)/(2)}}{y}

Thus, The simplified form of given expression\frac{15xy}{5x^{(1)/(2)}y^2} is \frac{3x^{(1)/(2)}}{y}

Answer:

3x^{(1)/(2)}y^(-1)

Step-by-step explanation:

The given expression is:

\frac{15xy}{5x^{(1)/(2)}y^2}

We have to simplify the above given expression.Thus,

Firstly, divide the constant terms, we get

(15)/(5)=3

Now, applying the exponent law, that is (x^a)/(x^b)=x^(a-b), we have

\frac{xy}{x^{(1)/(2)}y^2}=x^{1-(1)/(2)}y^(1-2)=x^{(1)/(2)}y^(-1)

Thus, the simplified form of the above given equation is:

3x^{(1)/(2)}y^(-1)