Solve for s.<br><br><br><br> 2/3s+5/6s=2/1<br> all the numbers are fractions

3. The diagonal of a rectangular field is 169 m. Ifthe ratio of the length to the width is 12:5, find the: (a) (b) dimensions; p

Kazeer [188]

The dimensions of the rectangle are length 156 m and a width of 65m, and a perimeter P = 442m

<h3>How to find the dimensions of the rectangle?</h3>

For a rectangle of length L and width W, the diagonal is:

$D = \sqrt{L^2 + W^2}$

Here we know that the diagonal is 169m.

And the ratio of the length to the width is 12:5

This means that:

W = (5/12)*L

Replacing all that in the diagonal equation:

$169 = \sqrt{((5/12)^2*L^2 + L^2} \\\\169^2 = (25/144)*L^2 + L^2 = ( 25/144 + 1)*L^2\\\\\frac{169}{\sqrt{25/144 + 1} } = L = 156$

So the length is 156 meters, and the width is:

W = (5/12)*156 m = 65m

Finally, the perimeter is:

P = 2*(L + W) = 2*(156 m + 65m) = 442m

If you want to learn more about rectangles:

brainly.com/question/17297081

#SPJ1

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2 years ago

Jeremy bought 3 pairs of pants for $68.25. Each pair of pants cost the same amount. How much will he pay for 5 pairs of pants at

melamori03 [73]

Answer:

113.75

Step-by-step explanation:

3 0

3 years ago

Terry is doing his homework he is trying to find the value of x:2 = 30:12 what is the value of x

Luba_88 [7]

jcp-fnoa-kzk join here

8 0

3 years ago

Read 2 more answers

The flower shop has 40 times as many flowers in the cooler as Kendall has in her bouquet. Kendall has 15 flowers. How many flowe

user100 [1]

Answer:

tu debes de saber las respuestas por qué tú estás como en onceavo

3 0

3 years ago

The box at the right is a cube with edges that measure 2 feet. The sides of the triangle inside the cube are a diagonal of a fac

zvonat [6]

Answer:

$Area = 2 \sqrt{2}$

Step-by-step explanation:

Given

$Length_1 = 2$

$Length_2 = \sqrt{2^3}$

Required

Determine the area of the triangle

The given lengths represent the height and base length of the triangle.

So, the area is;

$Area = \frac{1}{2} * Length_1 * Length_2$

Substitute values for Length1 and Length2

$Area = \frac{1}{2} * 2 * \sqrt{2^3}$

$Area = \frac{2}{2} * \sqrt{2^3}$

$Area = 1* \sqrt{2^3}$

$Area = \sqrt{2^3}$

Express 2^3 as 2 * 2 * 2

$Area = \sqrt{2*2*2}$

$Area = \sqrt{4*2}$

Split

$Area = \sqrt{4} *\sqrt{2}$

$Area = 2 *\sqrt{2}$

$Area = 2 \sqrt{2}$

6 0

3 years ago

Solve for s. 2/3s+5/6s=2/1 all the numbers are fractions