All categories

2 years ago

True or False? If a radius of a circle intersects a chord, it is perpendicular to that chord.

Mathematics

2 answers:

evablogger [386]2 years ago

5 0

Answer:

False

Step-by-step explanation:

If a radius of circle intersect a chord then it is not necessary that it bisect the chord.

If the radius bisects the chord then it will be perpendicular to chord.

But, if does not bisect the chord then it won't be perpendicular to that chord.

The radius is perpendicular to chord when it bisects that chord .

Hence, if a radius of a circle intersects a chord, it is not perpendicular to that chord.

Answer : False

kifflom [539]2 years ago

4 0

If a radius of a circle intersects a chord, it would be unlikely for it to be perpendicular to that chord, so your answer choice for this question should be false.

You might be interested in

On a multiple choice test, each question has 7 possible answers. If you make a random guess on the first question, what is the p

valkas [14]

Answer:

c (1/7)

Step-by-step explanation:

4 0

3 years ago

Write the rule for the function represented by this table.

sineoko [7]

Rule: y= 3 • x + 1

Explanation

You times a the number shy three then add one

4 0

1 year ago

4 (2x-6)=10x-6. solve for x

Reika [66]

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*(2*x-6)-(10*x-6)=0

2x - 6 = 2 • (x - 3)

8 • (x - 3) - (10x - 6) = 0

-2x - 18 = -2 • (x + 9)

-2 • (x + 9) = 0

Solve : -2 = 0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :
Solve : x+9 = 0

Subtract 9 from both sides of the equation :
 x = -9

7 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

Answer the question in the picture. If its right I’ll mark brainliest

Kisachek [45]

Answer:

The answer is c!

Step-by-step explanation:

5 0

3 years ago