What is the solution to= 6

Mathematics

2 answers:

Elan Coil [88]3 years ago

6 0

Answer:

Answer: A. x + (4x - 85) = 90

In this case, the two angles are complementary. This means they add up to 90°. Therefore, the equation is x + (4x - 85) = 90.

Step-by-step explanation:

Volgvan3 years ago

4 0

Answer: y=18

Step-by-step explanation:

(y/6) * (3) = (6) * (3) = y=18

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daser333 [38]

The answer is -1/2 :)))

5 0

3 years ago

Read 2 more answers

AB=diameter of the circle O. OC=radius. Arc AXB is an arc of the circle with centre C. Prove areas of the shaded region are =

alexandr1967 [171]

1.
Draw a circle with center C And radius CA, as shown in the attached picture.

Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|= $\sqrt{2}R$ .

these are all shown in the picture.

2.

Area of triangle ABC is 1/2 * 2R * R= R^2

3.

Let the area between arc BXA and chord AB be Y. (the yellow region).

and let G be the shaded region between arcs AB and AXB.

G=1/2(Area circle with center O)-Y
= $\frac{1}{2} \pi R^{2}-Y$

To find Y:

Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.

So Area of sector ACB = $\frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}$

Y =area of sector ABC-Area(triangle ABC)= $\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2}$

4.

Finally,

$G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}$

This proves that the 2 shaded regions have equal area.