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3 years ago

. If Triangle ABC is equilateral, solve for X... * 7 (8x - 44)

Mathematics

2 answers:

Tasya [4]3 years ago

5 0

If triangle ABC is equilateral,solve for X56x-308

ki77a [65]3 years ago

5 0

Answer:

x=13

Step-by-step explanation: (8x-44)+(8x-44)+(8x-44)=180

24x-132=180

24x=312

312/24

x=13

Showing Proof

8(13)-44=60

60+60+60=180

Therefore the answer is x=13

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Marrrta [24]

Given:

A figure of a circle and two secants on the circle from the outside of the circle.

To find:

The measure of angle KLM.

Solution:

According to the intersecting secant theorem, if two secant of a circle intersect each other outside the circle, then the angle formed on the intersection is half of the difference between the intercepted arcs.

Using intersecting secant theorem, we get

$\angle KLM=\dfrac{1}{2}(Arc(JON)-Arc(KM))$

$(3x-4)=\dfrac{1}{2}(271-(x+6))$

$(3x-4)=\dfrac{1}{2}(271-x-6)$

Multiply both sides by 2.

$6x-8=265-x$

Isolate the variable x.

$6x+x=265+8$

$7x=273$

Divide both sides by 7.

$x=\dfrac{273}{7}$

$x=39$

Now,

$\angle KLM=(3x-4)^\circ$

$\angle KLM=(3(39)-4)^\circ$

$\angle KLM=(117-4)^\circ$

$\angle KLM=113^\circ$

Therefore, the measure of angle KLM is 113 degrees.

5 0

3 years ago

A paper clip is 2.35 cm. How many meters long is the clip?

sineoko [7]

The answer is 0.0235

3 0

3 years ago

Read 2 more answers

PLEASE!!!!!!!The adjoining figure shows two circles with the same center. The

Nookie1986 [14]

<h3>Answer:</h3>

$\boxed{\boxed{\pink{\sf \leadsto Hence \ the \ area \ of \ shaded \ region \ is 264 cm^2}}}$

<h3>Step-by-step explanation:</h3>

Here , two circles are given which are concentric. The radius of larger circle is 10cm and that of smaller circle is 4cm . And we need to find thelarea of shaded region.

From the figure it's clear that the area of shaded region will be the difference of areas of two circles.

Let the,

Radius of smaller circle be r .
Radius of smaller circle be r .
Area of shaded region be $\bf Area_{shaded}$

$\bf \implies Area_{Shaded}= Area_{bigger}-Area_{smaller} \\\\\bf\implies Area_{Shaded} = \pi R^2 - \pi r^2 \\\\\bf\implies Area_{shaded} = \pi ( R^2-r^2) \\\\\bf\implies Area_{shaded} = \pi [ (10cm)^2 - (4cm)^2] \\\\\bf\implies Area_{shaded} = \pi [ 100cm^2-16cm^2] \\\\\bf\implies Area_{shaded} = \pi \times 84cm^2 \\\\\bf\implies Area_{shaded} = \dfrac{22}{7}\times 84cm^2 \\\\\bf\implies \boxed{\red{\bf Area_{shaded} = 264 cm^2 }}$