Which polynomials are in standard form

PLEASE HELP MEEE !!!!

labwork [276]

Given:

The figure of a trapezium.

To find:

The measures of missing angle $m\angle K, m\angle M$ .

Solution:

In the given figure $JK\parallel ML$ .

If a transversal line intersect the two parallel lines, then the same sides interior angles are supplementary angles.

$m\angle J+m\angle M=180^\circ$ (Supplementary angles)

$87^\circ+m\angle M=180^\circ$

$m\angle M=180^\circ-87^\circ$

$m\angle M=93^\circ$

Similarly,

$m\angle L+m\angle K=180^\circ$ (Supplementary angles)

$51^\circ+m\angle K=180^\circ$

$m\angle K=180^\circ-51^\circ$

$m\angle K=129^\circ$

Therefore, $m\angle M=93^\circ$ and $m\angle K=129^\circ$ .

4 0

3 years ago

How do i solve for y and x?

Akimi4 [234]

Answer:

You use the facts about a 30, 60, 90 triangle.

Step-by-step explanation:

The angles in a triangle add up to 180.

90+30+x=180

120+x=180

x=60

This is a 30-60-90 right triangle.

The side opposite the 30 degree angle is a.

The side opposite the 90 degree angle (hypotenuse) is 2a.

The side opposite the 60 degree angle is a $\sqrt{3}$

We know the side opposite the 90 degree angle.

2a = 12 $\sqrt{3}$

Divide by 2.

a=6 $\sqrt{3}$

a is the side opposite the 30 degree angle (y)

Because we know a, we can find a $\sqrt{3}$ .

6 $\sqrt{3 } (\sqrt{3})$

The square roots cancel, leaving 3.

6 times 3 is 18.

Therefore, the side opposite the 60 degree angle (x) is 18.

4 0

3 years ago

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Joel built a model town. He constructed three buildings that each had a volume of 18 cm cubed. Each building looked different. W

morpeh [17]

Answer:

The three buildings can look like following:

1. The building can look like a cube, with each side and height equal to 2.62 cm.

2. The building can look like a cuboid, with 3 cm length, 1 cm breadth and 6 cm height.

3. The building can look like a pyramid, with 3 cm length, 3 cm breadth and 6 cm height.

Step-by-step explanation:

The three possibilities of the appearance or shape of the building, with the volume equal to 18 cm³ are as follows:

1. CUBIC SHAPE:

The building can look like a cube with the length of each side and height given as:

Volume of Cube = 18 cm³ = L³

L = ∛18 cm³ = 2.62 cm

Thus, the building can look like a cube, with each side and height equal to 2.62 cm.

2. CUBOID SHAPE:

The building can also look like cuboid with the following dimensions.

Length = 3 cm

Breadth = 1 cm

Height = 6 cm

Thus, the volume becomes:

Volume of Cuboid = Length * Breadth * Height

Volume of Cuboid = 3 cm * 1 cm * 6 cm = 18 cm³

Thus, the building can look like a cuboid, with 3 cm length, 1 cm breadth and 6 cm height.

3. PYRAMID SHAPE:

The building can also look like pyramid with the following dimensions.

Length = 3 cm

Breadth = 3 cm

Height = 6 cm

Thus, the volume becomes:

Volume of Pyramid = (Length * Breadth * Height)/3

Volume of Cuboid = (3 cm * 3 cm * 6 cm)/3 = 18 cm³

Thus, the building can look like a pyramid, with 3 cm length, 3 cm breadth and 6 cm height.

5 0

3 years ago

Please Help What equation is graphed in this figure?

Airida [17]

Answer:

I think that is quadratic

4 0

3 years ago

I could really use some help on this one 1/2(b+14)=b+14÷2

konstantin123 [22]

B+14/2=b+14/2
1
they r equal

4 0

3 years ago

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