Alexa is designing a paper airplane whose final shape, when viewed from the top or bottom, is a trapezoid. a sketch of her plane, viewed from the top, is shown on the left.
what are the dimensions of one of the identical triangular pieces of the plane?
2 cm base, 3 cm height
3 cm base, 3 cm height
3 cm base, 4 cm height
3 cm base, 6 cm height

2 cm base, 3 cm height

Step-by-step explanation:

Let

x > the height of one of the identical triangles pieces of the plane

y > the base of one of the identical triangles pieces of the plane

we know that

x=3 cm

2y+2=6

solve for y

2y=6-2

2y=4

y=2 cm

therefore

The dimensions of one of the identical triangles pieces of the plane are 2 cm base and 3 cm height

The correct option is 1. The dimensions of one of the identical triangular pieces of the plane are 2 cm base and 3 cm height.

Step-by-step explanation:

From the given figure it is noticed that the length of parallel sides of the trapezoid are 6 cm and 2 cm.

The height of triangular pieces is same as the height of trapezoid. So the height of triangular pieces is 3 cm.

Let the measure of base of the triangular pieces be x.

Combine like terms.

Subtract 2 form both the sides.

Divide both sides by 2.

The base of the triangles is 2 cm.

The dimensions of one of the identical triangular pieces of the plane are 2 cm base and 3 cm height. Therefore the correct option is 1.

answer: 298143

step-by-step explanation:

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}