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1 year ago

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Points E, F, and D are located on circle C. Circle C is shown. Line segments E C and D F are radii. Lines are drawn from points

E and D to point F to form chords E F and D F. Arc E D is 68 degrees. The measure of arc ED is 68°. What is the measure of angle EFD? 34° 68° 112° 132°

2 answers:

elena55 [62]1 year ago

7 0

Answer:

Option (1). 34°

Step-by-step explanation:

From the figure attached, CE and CD are the radii of the circle C.

Central angle CED formed by the intercepted arc DE = 68°

Since measure of an arc = central angle formed by the intercepted arc

Therefore, m∠CED = 68°

Since m∠EFD = $\frac{1}{2}(m\angle ECD)$ [Central angle of an intercepted arc measure the double of the inscribed angle by the same arc]

Therefore, m∠EFD = $\frac{1}{2}(68)$

= 34°

Therefore, Option (1) 34° will be the answer.

marusya05 [52]1 year ago

7 0

Answer:

34 (100% Correct)

Step-by-step explanation:

You might be interested in

Thomas graphed the line that represents the equation y=34x.

zloy xaker [14]

Answer:

The ordered pairs represent points on the line are

(4, 3) ⇒ C

(2, $\frac{3}{2}$ ) ⇒ D

(-8, -6) ⇒ E

Step-by-step explanation:

To find the ordered pairs represent points on the line, substitute x by the x-coordinate of each point, if the value of y equals the y-coordinate of the point, then the point is on the line.

∵ The equation is y = $\frac{3}{4}$ x

∵ The ordered pair is (8, $\frac{1}{6}$ )

→ Substitute x by 8

∴ y = $\frac{3}{4}$ (8)

∴ y = 6

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair (8, $\frac{1}{6}$ ) does not represent a point on the line

∵ The ordered pair is ( $\frac{-2}{3}$ , $\frac{1}{2}$ )

→ Substitute x by $\frac{-2}{3}$

∴ y = $\frac{3}{4}$ ( $\frac{-2}{3}$ )

∴ y = $\frac{-1}{2}$

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair ( $\frac{-2}{3}$ , $\frac{1}{2}$ ) does not represent a point on the line

∵ The ordered pair is (4, 3 )

→ Substitute x by 4

∴ y = $\frac{3}{4}$ (4)

∴ y = 3

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (4, 3) represents a point on the line

∵ The ordered pair is (2, $\frac{3}{2}$ )

→ Substitute x by 2

∴ y = $\frac{3}{4}$ (2)

∴ y = $\frac{3}{2}$

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (2, $\frac{3}{2}$ ) represents a point on the line

∵ The ordered pair is (-8, -6 )

→ Substitute x by -8

∴ y = $\frac{3}{4}$ (-8)

∴ y = -6

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (-8, -6) represents a point on the line

5 0

1 year ago

Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right

Jet001 [13]

Answer:

Option B.

Step-by-step explanation:

It is given that ΔSRQ is a right angle triangle, ∠SRQ is right angle.

RT is altitude on side SQ, ST=9, TQ=16 and SR=x.

In ΔSRQ and ΔSTR,

$m\angle S=m\angle S$ (Reflexive property)

$m\angle R=m\angle T$ (Right angle)

By AA property of similarity,

$\triangle SRQ\sim \triangle STR$

Corresponding parts of similar triangles are proportional.

$\dfrac{SR}{SQ}=\dfrac{ST}{SR}$

Substitute the given values.

$\dfrac{x}{9+16}=\dfrac{9}{x}$

$\dfrac{x}{25}=\dfrac{9}{x}$

On cross multiplication we get

$x^2=25\times 9$

$x^2=225$

Taking square root on both sides.

$x=\sqrt{225}$

$x=15$

The value of x is 15. Therefore, the correct option is B.

7 0

2 years ago

Read 2 more answers

I need help ASAP!!!!!When bisecting an angle, three arcs are drawn. One is used to find points on the rays of the angle, and the

user100 [1]

Answer:

B

Step-by-step explanation:

The order does not matter. The intersection point will be the same regardless of the order the arcs are drawn in.

4 0

1 year ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

In a necklace the ratio of yellow beads to blue beads is 3:1. What percentage of beads are yellow?

navik [9.2K]

3 yellow beads and 1 blue bead

4 0

2 years ago

Read 2 more answers