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3 years ago

A circle has a diameter with endpoints at (0, 4) and (0, –8). What is the approximate area of the circle?

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

6 0

Answer:

$36\pi$

Step-by-step explanation:

The diameter is:

The Euclidean distance between points (x1, y1) and (x2, y2) is:

sqrt((x1 - x2)^2 + (y1 - y2)^2)

Substitute (x1, y1) = (0, 4) and (x2, y2) = (0, -8):

= sqrt((-0 + 0)^2 + (-(-8) + 4)^2)

= sqrt(12^2)

= 12

or simply:

diameter: 4-(-8)=12

The radius is:

12/2=6 units

The area of the circle:

$\pi r^2=\pi(6^2)=36\pi$

nadya68 [22]3 years ago

4 0

Answer:

113.04

Step-by-step explanation:

Area of a circle is πr2

π = 3.14

r = 6

πr2 = 113.04

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3(4x−2)=2(5x+3) what is the answer pls asap

Paha777 [63]

Answer:

Step-by-step explanation:

1. get rid of parentheses: 12x -6 = 10x +6

2. group like terns together: 12x-10x = 6+6

3. simplify: 2x = 12

4. divide both sides by 2: x=6

7 0

3 years ago

Read 2 more answers

To graph the equation 2x + 5y = 10, Zeplyn draws a line through the points (5, 0) and (0, 2). What is the slope of the line repr

BARSIC [14]

Hi there! The slope of the line is -2/5.

We can determine the slope of a line by dividing the change of y-coordinates by the change of x-coordinates.

In this situation we have a...
change of x-coordinates of (5 - 0) = 5
change of y-coordinates of (0 - 2) = -2

Therefore, the slope of this line is -2/5

6 0

3 years ago

Martin bikes at a constant speed to a finish line that is 15 mi away. After 20 mins, he is 10 mi from the finish line. After 30

ankoles [38]

The answer is C. -0.25; the rate at which Martin's distance from the finish line changes per minute

5 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

A.) it has no reflectional symmetry

coldgirl [10]

Answer:

D.) It has rotational symmetry with an angle of rotation of 90 degrees.

Step-by-step explanation:

i) the shape shown in the figure has reflectional symmetry with four lines of symmetry. There are four lines of symmetry such that each line when drawn divides the shape into two exact halves such that if each half were too be folded across the line of symmetry it would produce the other half exactly.

Therefore Option A.) and Option B.) are not true

ii) It has point symmetry which means if the shape is rotated by 180 degrees the shape does not change. Therefore Option C.) is not true.

iii) the shape has rotational symmetry because if the shape is rotated though any between 0 degrees and 360 degrees the shape appears to be unchanged.

Therefore the correct options are

D.) It has rotational symmetry with an angle of rotation of 90 degrees.

5 0

3 years ago