The 'diameter' is the line segment that passes through the center of the circle.
In the circle shown, F is the center of the circle.
The line segments EB and AC passes through the center point F.
Therefore, both EB and AC are the diameters of the given circle.
From the options given, option c) AC is the correct answer.

5 0

4 years ago

Given the function g(x) = 6^x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

Ad libitum [116K]

Section A and B
$\frac{36 - 6}{2 - 1} = 30 \\ \frac{1296 - 216}{4 - 3} = 1080$

7 0

4 years ago

Given: KLMN is a trapezoid, m∠K = 90°

valentina_108 [34]

A=162sq......... bless up

5 0

3 years ago

Read 2 more answers

1) Given: circle k(O), ED= diameter ,m∠OEF=32°, m(arc)EF=(2x+10)° Find: x

qaws [65]

1. The major arc ED has measure 180 degrees since ED is a diameter of the circle. The measure of arc EF is $(2x+10)^\circ$ , so the measure of arc DF is

$m\widehat{DF}=360^\circ-180^\circ-(2x+10)^\circ=(170-2x)^\circ$

The inscribed angle theorem tells us that the central angle subtended by arc DF, $\angle DOF$ , has a measure of twice the measure of the inscribed angle DEF (which is the same angle OEF) so

$m\angle DOF=2m\angle OEF=64^\circ$

so the measure of arc DF is also 64 degrees. So we have

$170-2x=64\implies106=2x\implies\boxed{x=53}$

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2. Arc FE and angle EOF have the same measure, 56 degrees. By the inscribed angle theorem,

$m\angle EOF=2m\angle EDF\implies56^\circ=2m\angle EDF\implies m\angle EDF=28^\circ$

Triangle DEF is isosceles because FD and ED have the same length, so angles EFD and DEF are congruent. Also, the sum of the interior angles of any triangle is 180 degrees. It follows that

$m\angle EFD+m\angle EDF+m\angle DEF=180^\circ\implies\boxed{m\angle EFD=76^\circ}$

Triangle OFE is also isosceles, so angles EFO and FEO are congruent. So we have

$m\angle EFO+m\angle FEO+m\angle EOF=180^\circ\implies\boxed{m\angle EFO=62^\circ}$

7 0

3 years ago