All categories

3 years ago

What is the center of the circle that has a diameter whose endpoints are (6, -4) and (18, 10)?

Mathematics

2 answers:

Maurinko [17]3 years ago

8 0

Answer:

(12, 3)

Step-by-step explanation:

To answer this, we find the midpoint of the line segment (that is, of the diameter) connecting (6, -4) and (18, 10):

The formula for the midpoint gives us x-midpoint = (6+18)/2, or x-midpoint = 12.

The formula for the midpoint gives us y-midpoint = (-4+10)/2, or x-midpoint = (6)/2 = 3.

Thus, the center of this circle is at (x-midpoint), y-midpoint), or (12, 3).

The third answer choice is the correct one.

Leno4ka [110]3 years ago

7 0

Answer:

(12, 3)

Step-by-step explanation:

You might be interested in

Which expression is represented by this model?

Sophie [7]

Answer:

the first one

Step-by-step explanation:

hope i helped

8 0

3 years ago

Sorry it’s small I could not get it in just zoom in

Ne4ueva [31]

Hello!

Graphs A and D are reflected, not rotated, so they are not the answer.

Graph C is rotated 180°, so it is not the answer.

This leaves us with B as our final answer.

I hope this helps!

3 0

3 years ago

What is 395.20 divided by 5 rounded

olchik [2.2K]

I think it is 395.20 / 5 = 79.04, rounded down to 79

3 0

3 years ago

Read 2 more answers

You have constructed wooden forms for pilings. The pilings are in the shape of rectangular solids, each measuring 10 ft. high an

Anna [14]

Volume V=3*10=30 feet³ -- > one piling

10 pilings----------------- > V=30*10=300 feet³

The answer is 300 feet³ of cement

3 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