If one angle in a parallelogram is twice the measure of the consecutive angle, what are the measures of the angles?
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The measure of the angles of a parallelogram are ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°.

Step-by-step explanation:

The consecutive angles of a parallelogram are supplementary, i.e. they sum up to 180°.

Consider the parallelogram ABCD.

∠A and ∠D and ∠B and ∠C are consecutive angles.

In a parallelogram opposite angles are equal.

That is: ∠A = ∠C and ∠B = ∠D.

Using the provided information, suppose ∠A = 2 × ∠D.

Compute the value of ∠A and ∠D as follows:

∠A + ∠D = 180°

2∠D + ∠D = 180°

3∠D = 180°

∠D = 60°

Then the measure of ∠A is:

∠A = 2∠D

     = 2 × 60°

     = 120°

Now the measure of ∠B and ∠C are:

∠C = ∠A = 120°

∠B = ∠D = 60°

Thus, the measure of the angles of a parallelogram are ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°.

the measures of the angles are 120 degrees and 60 degrees.

Step-by-step explanation:

In a parallelogram, the opposite angles are equal and the consecutive angles are supplementary. This means that the sum of the angles is 180 degrees.

Let x represent the measure of one of the angles of the parallelogram. The consecutive angle would be 180 - x

If one angle in a parallelogram is twice the measure of the consecutive angle, it means that

x = 2(180 - x)

x = 360 - 2x

2x + x = 360

3x = 360

x = 360/3

x = 120 degrees

The other angle is

180 - 120 = 60 degrees

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}