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3 years ago

A circle has an area of 324╥ cm². What is the radius?

2 answers:

8 0

The area of a circle has the formula A = pi * r^2. Since the area is known, we could determine the radius of the circle.

A = pi*r^2
324*pi = pi*r^2
324 = r^2
r = square root (324)
r = 18

The answer is 18 cm.

Stolb23 [73]3 years ago

3 0

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Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

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3 years ago

What is the domain of the relation y = arccscx?

Stella [2.4K]

Answer:

$y=arc\csc x$

Step-by-step explanation:

The given function is $y=arc\csc x$ .

The domain refers to all values of x for which this function is defined.

Recall that: the domain of $y=arc \sin (x)$ is $-1\le x\le1$

And we know $y=arc\csc x$ is the reciprocal of $y=arc \sin (x)$ .

Therefore the complement of the domain of $y=arc \sin (x)$ which is $(-\infty,-1]\cup [1,+\infty)$ is the domain of $y=arc\csc x$

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3 years ago