Open notes test: Relations, Functions and Slope

Help on #6 please i really need help

Wewaii [24]

Answer:

decreased 11.7647 percent

5 0

3 years ago

Please answer the 3 questions (the first one I accidentally tapped)

choli [55]

1,347 = 1.347 x 10^3

2,156 = 2.156 x 10^3

9,200,000 = 9.2 x 10^6

8 0

3 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 923.558 m²

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, $A_{req}$ ≈ 923.558 m²

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

3 years ago

Theorem 8.18 and the Pythagorean Theorem to find the side lengths of the kite . Write the lengths in simplest radical form and a

Viktor [21]

Answer:

$|WX|=|XY|=\sqrt{18} =3\sqrt{2}\:Units \:or\: 4.2 \:Units\\|WZ|=|YZ|=\sqrt{34} \:Units \:or\: 5.8 \:Units$

Step-by-step explanation:

Theorem: The diagonals of a kite are perpendicular.

Let O be the point of intersection of the diagonals,

Applying Pythagoras Theorem, in right triangle WOX

$|WX|^2=|WO|^2+|OX|^2\\|WX|^2=3^2+3^2=18\\|WX|=\sqrt{18} =3\sqrt{2}\:Units \:or\: 4.2 \:Units$

Applying Pythagoras Theorem, in right triangle WOZ

$|WZ|^2=|WO|^2+|OZ|^2\\|WX|^2=3^2+5^2=34\\|WZ|=\sqrt{34} \:Units \:or\: 5.8 \:Units$

4 0

3 years ago

How to show the tens fact you used. write the difference. 12-7=____ 10-___=____

prisoha [69]

The first answer is 5,but i don't know about the second one. (sorry)

3 0

3 years ago