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Alona [7]
3 years ago
10

Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

D=6\ cm

The radius of circle A is

r=6/2=3\ cm -----> the radius is half the diameter

<em>Circle B</em>

r=18\ cm

Compare the radius of both circles

3\ cm< 18\ cm

18=6(3)

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

6^{2}=36

therefore

The area of circle B is 36 times greater than the area of circle A

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PLEASE HELP ME.<br> THANKS!!!
Veronika [31]

Cos(x) = sin(x + 20o) Use the double cos angle formula

Cos(x) = sin(x)*cos(20) + cos(x)*sin(20) Divide through by cos(x). Tanx = sin(x)/cos(x)

1 = tan(x)*cos(20) + sin(20) Subtract sin(20) from both sides.

1 - sin(20) = tan(x)*cos(20) Calculate the value of 1 - sin(20)

0.65798 = tan(x) * 0.939692 Divide by cos(20)

tan(x) = 0.65795/0.939692

tan(x) = 0.70021 Take the inverse of 0.7) = tan(x)

x = tan-1(0.70021)

x = 35 degrees as expected.

Problem Two

The easiest way to do this is to pick random values and try it. The two acute angles are complementary. So 25 and 65 are random enough.

Let A = 25

Let C = 65

A

Tan(25) = sin(25)/sin(65) ; tan(25) = 0.4663.

sin(25)/sin(65) = 0.4663 Answer

B

Sin(90 - C) = Sin(A)

Tan(90 - A) = Tan(C)

So the question becomes Cos(A) = Tan(C) / sin(A)

Cos(25) = Tan(65)/Sin(25)

0.906 = ? 5.07 This statement isn't true.

C

Sin(65) = Cos(25)/Tan(65)

.906 = 0.4226 C is not the correct answer.

D

D isn't true. The tan does not relate that away. You can find it for yourself.

Cos(A) = Tan(C) You should get 0.906 = 2.14

E

Sin(C) = Cos(A) / Tan(A) I'll leave you to show this is wrong.

Problem 3

The diagram below is for this problem. Cos(x) = 50/100 = 0.5

x = cos-1(0.5)

x = 60 degrees.

Part 2

Sin(60) = opposite / hypotenuse

opposite = sin(60) * hypotenuse

opposite = 86.61 Be sure and round this to whatever the question says to round it to.



4 0
3 years ago
How to solve 6x-3=7x-11
VladimirAG [237]
Hello there.

<span>How to solve 6x-3=7x-11</span>
<span><span><span>6x</span>−3</span>=<span><span>7x</span>−11</span></span>
<span><span><span><span>6x</span>−3</span>−<span>7x</span></span>=<span><span><span>7x</span>−11</span>−<span>7x</span></span></span><span><span><span>−x</span>−3</span>=<span>−11</span></span>
<span><span><span><span>−x</span>−3</span>+3</span>=<span><span>−11</span>+3</span></span><span><span>−x</span>=<span>−8</span></span>
<span><span><span>−x/</span><span>−1</span></span>=<span><span>−8/</span><span>−1</span></span></span><span>x=8</span>
Answer: 8
6 0
3 years ago
How many triangles can be made if one angle is 95° and another angle is acute?
Oksana_A [137]
Theoretically, the answer will be infinite.

Since one angle of 95 degrees means the other two angles must be acute. You can pick any two real numbers with the sum of 85.
8 0
3 years ago
What is the cross section formed by a plane that intersects three faces of a cube
dusya [7]
Imagine taking a cube and slicing it open from one of the corners.

The exposed area of the cube would be a triangle. See the image for reference.

7 0
3 years ago
An elliptical-shaped path surrounds a garden, modeled by quantity x minus 20 end quantity squared over 169 plus quantity y minus
Margaret [11]

The <em>maximum</em> distance between any two points of the ellipse is 34 feet.

<h3>Procedure - Determination of the distance between two points of a ellipse</h3>

The maximum distance between any two points of a ellipse is the <em>maximum</em> distance between the ends of the ellipse along the <em>longest</em> axis, which is parallel to the y-axis in this case.

In addition, the equation of the ellipse in <em>standard</em> form is defined by this formula:

\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 (1)

Where:

  • h, k - Coordinates of the center of the ellipse.
  • a, b - Lengths of each semiaxis.

Hence, the maximum distance (d_{max}), in feet, is calculated by this formula:

d_{max} = 2\cdot b (2)

If we know that b = 17, then the maximum distance between any two points of the ellipse is:

d_{max} = 2\cdot (17\,ft)

d_{max} = 34\,ft

The <em>maximum</em> distance between any two points of the ellipse is 34 feet. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted, correct form is presented below:

<em>An elliptical-shaped path surrounds a garden, modeled by </em>\frac{(x-20)^{2}}{169} + \frac{(y-18)^{2}}{289} = 1<em>, where all measurements are in feet. What is the maximum distance between any two points of the path.</em>

To learn more on ellipses, we kindly invite to check this verified question: brainly.com/question/19507943

5 0
2 years ago
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