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3 years ago

A rectangular field is 0.3 kilometers long and 0.15 kilometers wide. What is the area of the field in square meters?

Mathematics

1 answer:

Anika [276]3 years ago

6 0

Answer:

4.5 meters

Step-by-step explanation:

The formula for the area of a rectangle is length x width. Using this formula, you can substitute the given information and find the area:

length x width = area

0.3km x 0.15km = area

0.045km = area

Finally, you can now convert square kilometres into square meters by multiplying by 100:

0.045 x 100 = 4.5 meters

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The volume of a box(V) varies directly with its length(l). If one of the boxes has a volume of 288 cubic inches and a length of

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Step-by-step explanation:

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3 years ago

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The first pair of statements is the best option. h(0) based on the given equation would be the square root of -3, which is an imaginary number. Also, w(0) should be 0, since if nobody is working on making wrenches, none can be made. This is despite the graph having a value at x = 0.

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4 years ago

What is x if 2x+7 + 13 is equal to 3x

Anastasy [175]

Answer:

x = 20

Step-by-step explanation:

2x+7+13 = 3x

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4 years ago

Read 2 more answers

A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately <u>923.558 m²</u>

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

3 years ago