How do you complete the other two? I know how to complete the first one but 3D Pythag confuses me so much

For now, I'll focus on the figure in the bottom left.

Mark the point E at the base of the dashed line. So point E is on segment AB.

If you apply the pythagorean theorem for triangle ABC, you'll find that the hypotenuse is

a^2+b^2 = c^2

c = sqrt(a^2+b^2)

c = sqrt((8.4)^2+(8.4)^2)

c = 11.879393923934

which is approximate. Squaring both sides gets us to

c^2 = 141.12

So we know that AB = 11.879393923934 approximately which leads to (AB)^2 = 141.12

Now focus on triangle CEB. This is a right triangle with legs CE and EB, and hypotenuse CB.

EB is half that of AB, so EB is roughly AB/2 = (11.879393923934)/2 = 5.939696961967 units long. This squares to 35.28

In short, (EB)^2 = 35.28 exactly. Also, (CB)^2 = (8.4)^2 = 70.56

Applying another round of pythagorean theorem gets us

a^2+b^2 = c^2

b = sqrt(c^2 - a^2)

CE = sqrt( (CB)^2 - (EB)^2 )

CE = sqrt( 70.56 - 35.28 )

CE = 5.939696961967

It turns out that CE and EB are the same length, ie triangle CEB is isosceles. This is because triangle ABC isosceles as well.

Notice how CB = CE*sqrt(2) and how CB = EB*sqrt(2)

Now let's focus on triangle CED

We just found that CE = 5.939696961967 is one of the legs. We know that CD = 8.4 based on what the diagram says.

We'll use the pythagorean theorem once more

c = sqrt(a^2 + b^2)

ED = sqrt( (CE)^2 + (CD)^2 )

ED = sqrt( 35.28 + 70.56 )

ED = 10.2878569196893

This rounds to 10.3 when rounding to one decimal place (aka nearest tenth).

Now I'm moving onto the figure in the bottom right corner.

Draw a segment connecting B to D. Focus on triangle BCD.

We have the two legs BC = 3.7 and CD = 3.7, and we need to find the length of the hypotenuse BD.

Like before, we'll turn to the pythagorean theorem.

a^2 + b^2 = c^2

c = sqrt( a^2 + b^2 )

BD = sqrt( (BC)^2 + (CD)^2 )

BD = sqrt( (3.7)^2 + (3.7)^2 )

BD = 5.23259018078046

Which is approximate. If we squared both sides, then we would get (BD)^2 = 27.38 which will be useful in the next round of pythagorean theorem as discussed below. This time however, we'll focus on triangle BDE

a^2 + b^2 = c^2

b = sqrt( c^2 - a^2 )

ED = sqrt( (EB)^2 - (BD)^2 )

x = sqrt( (5.9)^2 - (5.23259018078046)^2 )

x = sqrt( 34.81 - 27.38 )

x = sqrt( 7.43 )

x = 2.7258026340878

x = 2.7

As an alternative, we could use the 3D version of the pythagorean theorem (similar to what you did in the first problem in the upper left corner)

The 3D version of the pythagorean theorem is

a^2 + b^2 + c^2 = d^2

where a,b,c are the sides of the 3D block and d is the length of the diagonal. In this case, a = 3.7, b = 3.7, c = x, d = 5.9

So we get the following

a^2 + b^2 + c^2 = d^2

c = sqrt( d^2 - a^2 - b^2 )

x = sqrt( (5.9)^2 - (3.7)^2 - (3.7)^2 )

x = 2.7258026340878

x = 2.7

Either way, we get the same result as before. While this method is shorter, I think it's not as convincing to see why it works compared to breaking it down as done in the previous section.

when studying polynomials, you often hear the terms zeros, roots, factors and x x the multiplicity of a zero is important because it tells us how the graph of the let's look at the graph of a function that has the same zeros

step-by-step explanation: