There isn't enough info to prove the triangles to be congruent or not. So we can't say for sure either way.
We have angle CAD = angle ACB given by the arc markings, and we know that AC = AC due to the reflexive theorem. However we are missing one third piece of information.
That third piece of info could be....
- AD = BC which allows us to use SAS
- angle ACD = angle CAB which allows us to use ASA
- angle ABC = angle CDA which allows us to use AAS (slight variation of ASA)
Since we don't know any of those three facts, we simply don't have enough information.
side note: If AB = CD, then this leads to SSA which is not a valid congruence theorem. If we had two congruent sides, the angle must be between the two sides, which is what AD = BC allows.
Answer:
4.13=4.130
If there is a 0 at the end of a decimal then you can drop it.
This is a problem of maxima and minima using derivative.
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =

Surface area of the rectangular sides =

Therefore, the total area of the cube is:

Isolating the variable y in terms of x:

Substituting this value in V:

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

Solving for x:

Solving for y:

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ftWidth = 4 ftHeight = 2 ftAnd its volume is:
Answer:
= -30x -36
Step-by-step explanation:
-3x . 2-24x - 36= -30x - 36
-3x . 2 -24x - 36
= -6x - 24x - 36
= -30x - 36