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Liula [17]
3 years ago
8

The circle below is centered at the origin and has a radius of 2. What is its equation?

Mathematics
2 answers:
Elan Coil [88]3 years ago
7 0
C . X^2 + y^2 = 2^2= 4
Vitek1552 [10]3 years ago
3 0
Hello!

The equation for a circle is

(x - h)^2 + (y - k)^2 = r^2

(h, k) is the center
r is the radius

Put in the values you know

(x - 0)^2 + (y - 0)^2 = 2^2

Simplify

x^2 + y^2 = 4

The answer is C) x^2 + y^2 = 4

Hope this helps!
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Find the fifth roots of 243(cos 240° + i sin 240°).
Mazyrski [523]

Answer:

See below.

Step-by-step explanation:

Fifth root of 243 = 3,

Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),

then r^5( cos  Ф  + i sin  Ф )^5 = 243(cos 240 + i sin 240).

Equating equal parts and using de Moivre's theorem:

r^5 =243  and  cos  5Ф  + i sin  5Ф = cos 240 + i sin 240

r = 3 and  5Ф = 240 +360p so Ф =  48 + 72p

So Ф = 48, 120, 192, 264, 336  for   48 ≤ Ф < 360

So there are 5 distinct solutions given by:

3(cos 48 + i sin 48),

3(cos 120 + i sin 120),

3(cos 192 + i sin 192),

3(cos 264 + i sin 264),

3(cos 336 + i sin 336).. (Answer).

7 0
3 years ago
SOLVE
Varvara68 [4.7K]

Answer:

26.4 is what i came up with if you divide im probably wrong sorry if i am

5 0
3 years ago
Find the total area of the figure. It is _______ square inches.
kompoz [17]

Answer:

1100 square inches

Step-by-step explanation:

Formula: LxW

<u>Smallest rectangles</u> area: 5*20=100= 1 rectangle

there are 2 smal  rectangles so= 100*2=200

<u>Medium rectangles</u> area: 5*30=150

There are 2 so= 150*2= 300

<u>Biggest rectangle:</u> 30*20= 600

Total Area: 200+300+600= 1100 square inches

6 0
3 years ago
Twoooooooooooooooooooooooooooo
KatRina [158]

Answer:

n = 60.22

Step-by-step explanation:

Hello

To find Sn, we need to draw out equations for each a₇ and a₁₉

In an arithmetic progression,

Sn = a + (n-1)d

Where Sn = sum of the A.P

a = first term

d = common difference

a₇ = 32

32 = a + (7-1)d

32 = a + 6d ........equation (i)

a₁₉ = 140

140 = a + (19-1)d

140 = a + 18d .........equation (ii)

Solve equation (i) and (ii) simultaneously

From equation (i)

32 = a + 6d

Make a the subject of formula

a = 32 - 6d .....equation (iii)

Put equation (iii) into equation (ii)

140 = (32 - 6d) + 18d

140 = 32 - 6d + 18d

Collect like terms

140 - 32 = 12d

12d = 108

d = 108 / 12

d = 9

Put d = 9 in equation (i)

32 = a + 6(9)

32 = a + 54

a = 32 - 54

a = -22

When Sn = 511

Sn = a + (n - 1)d

Substitute and solve for n

511 = -22 + (n-1) × 9

511 = -22 + 9n - 9

511 = -31 + 9n

511 + 31 = 9n

542 = 9n

n = 542 / 9

n = 60.22

5 0
3 years ago
Please help i dont understand thank you
Alenkasestr [34]

Answer:

Part a)

We need to find the equation of a straight line passing through two given points in slope-intercept form

Part b)

The information given; we are given two points where the line passes through; (0, -4) and (-2, 2)

Part c)

We shall first determine the slope of the line using the formula;

change in y/change in x. Next, we determine the value of the y-intercept using the general form of the equation of a straight line in slope-intercept form; y = mx+c

Part d)

The slope of the line is calculated as;

(2--4)/(-2-0) =6/-2 = -3

The equation of the line in slope-intercept form becomes;

y = -3x +c

We use the point (0, -4) to determine the value of c;

-4 = -3(0)+c

c = -4

Part e)

Final solution thus becomes;

y=-3x-4

8 0
3 years ago
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