A wheel has a diameter of 28 inches. What is the approximate distance around the outside of the wheel? Use 3.14 for pi.

Order from least to greatest 2.34,2 3/4,9/4.

Paladinen [302]

Its not in order its 2.34,9/4,2 3/4

4 0

2 years ago

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

2 years ago

What is the value of sin 0 given that (3-4) is a point on the terminal side of 0

Effectus [21]

Using trigonometric relations, we will see that:

sin(θ ) = -0.8

<h3>How to get the sine of the angle?</h3>

For any point (x, y), the angle between the positive x-axis and the ray that connects the origin with the point (x, y) is:

θ = Atan(y/x)

In this case, the point is (3, -4), then the angle is:

θ = Atan(-4/3) = -53.13°

Then the sine of theta is:

sin(θ = -53.13°) = -0.8

If you want to learn more about trigonometric functions:

brainly.com/question/8120556

#SPJ1

8 0

1 year ago

Suppose j varies jointly with g and v, and j=1 when g=4 and v=5. what is j when g=10 and v=9?

Tems11 [23]

J $\alpha gv$
j=agv where a is a constant of proportionality.
j=1 when g=4 and v=5
1=a*4*5
1=20a
a=1/20
a= 0.05
j=0.05gv
When g=10 and v=9,
j=0.05*10*9
j=0.5*9
j=4.5

4 0

3 years ago

Plz help me so confused

Alex Ar [27]

If you look at the chart (about the chores) you can find out everthing

4 0

3 years ago