First, we draw our line.
|------------------------------------------------------------------------------------|
a e
Next, break up this line into segments using the information.
|----------------------|----------------------|--------------------|------------------|
a b c d e
The entire line is 29.
ab + bc + cd + de = ae
ab + bc + cd + de = 29
You also know that
bd = bc + cd
Due to midpoint theorem,
ab = bc
cd = de
Then,
2ab + 2cd = 29
The equations we will use are
bd = bc + cd eq1
2bc + 2cd = 29 eq2
Dividing both sides of the equation in eq2 yields
bc + cd = 14.5
bd = bc + cd
bd = 14.5
Slope intercept form follows this general equation...

With this information you have posted, all you can do is find slope, your m. To calculate for slope, you need to use this following equation...

Then, you pick one of your coordinate pairs to be 1 and the other to be 2. I will choose the first coordinate pair as 1, so...

You plug in -1 into your slope formula...

Now, to find your y-intercept, or b, you must plug in one of your coordinate pairs. In that equation, you may plug in the values (3,3), so you'll have...

Just solve for b and you will be done
Answer:
100 in²
Step-by-step explanation:
The area of the banner is equal to the area of the initial rectangle minus the area of the cutout triangle.
The rectangle has a height of 8 inches and width of 14 inches, so its area is:
A = (8 in) (14 in) = 112 in²
The triangle has a base of 8 inches and a height of 3 inches, so its area is:
A = ½ (8 in) (3 in) = 12 in²
So the area of the banner is 112 in² − 12 in² = 100 in².
we know that
<u>The Side-Splitter Theorem</u>: States that If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides proportionally
so
in this problem

therefore
<u>the answer is</u>
The segment length is GJ
Answer:
The larger cross section is 24 meters away from the apex.
Step-by-step explanation:
The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.
The length of the side of the hexagon is equal to the radius of the circle that inscribes it.
The area is

Where
is the radius of the inscribing circle (or the length of side of the hexagon).
Now we are given the areas of the two cross sections of the right hexagonal pyramid:
From these areas we find the radius of the hexagons:
Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that
form similar triangles with length
Therefore we have:

We put in the numerical values of
,
and solve for
:
