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2 years ago

Which equation could be used to solve the following problem?

Mathematics

1 answer:

-BARSIC- [3]2 years ago

4 0

115 +10p =305

Step-by-step explanation:

given,

let p day it takes

So,115+10p =305

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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

You are a plumber, and you are repairing a toilet leak. The job should take 0.9 hours. About how many minutes will it

monitta

Answer:

Step-by-step explanation:

multiply 60 by 0.9 and you get 54..?

6 0

3 years ago

Read 2 more answers

The value of a house is $165,000 and it loses 4% of its value every year.

astraxan [27]

1)Y=165000(1-0.04)^t
2)y=165000(1-0.04)^20=?? Use the calculator

8 0

3 years ago

Which ordered pair is part of the solution set of the system of inequalities graphed below?

Serjik [45]

They didn't make it easy. Grid lines are apparently 3 apart, but the offered coordinates are all multiples of 4. It appears the only point that is in the doubly-shaded area is ...

... B (-4, -10)

4 0

3 years ago

GIVING AWAY 10 POINTS!!!!!! What is the solution to the system of equations? x = y + 4

slega [8]

Answer:

x= -6, y= 10

Step-by-step explanation:

x= y + 4

2x = 3y - 2

If we replace "x" with "y+4" in the second equation we get:

2(y+4) = 3y - 2

2y + 8 = 3y - 2

2y - 3y = -2-8

-y = -10

y = 10

Now we can go back to the first equation and solve for x:

x = y + 4

x = 10 + 4 = 14

8 0

3 years ago