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Tpy6a [65]
3 years ago
5

A soccer field is 90 meters wide and 120 meters long.The coach asked the players to run from corner to corner diagonally across.

What is the diagonal distance?

Mathematics
1 answer:
Marat540 [252]3 years ago
5 0

The length of diagonal is 150 meters

<em><u>Solution:</u></em>

Given that soccer field is 90 meters wide and 120 meters long

The coach asked the players to run from corner to corner diagonally across

To find: length of diagonal

The figure is attached below

Let ABCD be a rectangle with BD as diagonal

From given,

length = DC = 120 meters

width = BC = 90 meters

The diagonal and two sides of rectangle forms a right angled triangle with diagonal as hypotenuse

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

By above definition, for right angled triangle DCB

BD^2 = BC^2 + DC^2

BD^2 = 90^2 + 120^2\\\\BD^2 = 8100 + 14400\\\\BD^2 = 22500

Taking square root on both sides,

BD = \sqrt{22500}\\\\BD = 150

Thus length of diagonal is 150 meters

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70cm

Step-by-step explanation:

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Since, both of the tank posses the same quantity of water then there volume is the same thing

For rectangular base

Volume= (Lenght × breadth × hheight)

If we substitute values

Lenght= 80 cm

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height= 45 cm

Volume= 80cm × 70cm × 45cm

Volume= 252000cm^3

For square base

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side =60cm.

But volume of square base= volume of

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252000cm^3= 60^2 × height

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3 years ago
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Step-by-step explanation:

(-4 , -10) ; (-4 , -4)

Distance = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}

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Hamburgers come packed 12 in a package. Hamburger buns come packed 10 in a package. If we want one hamburger for each bun for a
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5 hamburger packages and 6 bun packages

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Answer:

y=-\frac{158}{579}

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

Find the value of the determinant of each matrix, and divide:

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y=-\frac{158}{579}

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3 years ago
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