A radius of a circle is perpendicular to a chord if and only if it ________

Mathematics

2 answers:

tiny-mole [99]3 years ago

7 0

A radius of a circle is perpendicular to a chord if and only if it bisect the chord.

BigorU [14]3 years ago

4 0

There could be many different correct answers, such as
- A radius of a circle is perpendicular to a chord if and only if it (bisects the minor arc)
- A radius of a circle is perpendicular to a chord if and only if it (bisects the major arc)
- A radius of a circle is perpendicular to a chord if and only if it (bisects the subtended angle of the chord)
- A radius of a circle is perpendicular to a chord if and only if it (bisects the chord)

There may be other ones!

You might be interested in

9a - 3 (a - 6) = -6 a=___

leonid [27]

9a - 3(a - 6) = -6...distribute thru the parenthesis
9a - 3a + 18 = -6....subtract 18 from both sides
9a - 3a = -6 - 18 ...combine like terms
6a = - 24...divide both sides by 6
a = -24/6
a = - 4

7 0

2 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is a separable differential equation. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$