Pls help me with this.

Explanation:

Focus on triangle HKL. This is a right triangle with the horizontal leg KL = 2. The vertical leg is the unknown side HK. The hypotenuse is HL shown by the single tickmark.

Triangle HKJ is also a right triangle. It shares the leg HK and the hypotenuse JH is congruent to HL as indicated by the single tickmark.

By the hypotenuse leg theorem, we can show that triangle HKL is congruent to triangle HKJ

From there, the corresponding pieces JK and KL must be the same length. So we can say JK = 2 because KL = 2.

Now use the segment addition postulate

GK = GJ + JK

GK = 7 + 2

GK = 9

Focus now on triangle GKH and triangle MKH. We can prove these are congruent triangles using the hypotenuse leg theorem. Here HK = HK as mentioned before; also, GH = HM due to the double tickmarks.

As you can probably guess by now, the hypotenuse leg theorem only applies to right triangles.

Since  triangle GKH is congruent to triangle MKH, we know that

GK = KM

Since GK = 9, this makes KM = 9.

Therefore,

GM = GK + KM

GM = 9 + 9

GM = 18

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180