Answer:
If you want to know what b is then it is 16
Step-by-step explanation:
38 + b = 54
54- 38 = 16
38 + 16= 54
We have that
scale factor=3
we know that
[volume new cube]=[scale factor]³*[volume original cube]
[volume new cube]=[3]³*[volume original cube]-----> 27*[volume original cube]
the answer is
<span>the volume increases by a factor of 27</span>
Answer:
Up on the left, up on the right
Step-by-step explanation:
The given function is:

The degree of f(x) is 4 i.e. an even degree and the leading coefficient i.e. coefficient with highest powered variable is positive in sign.
The graph of a function with even degree always open on same side from both ends. This depends on the sign of leading coefficient what will be the direction of both ends. The positive sign indicates upward opening and negative sign indicates downward opening.
Since, the leading coefficient of f(x) is positive, it will open towards up from both right and left side. So, the correct option is the fourth option.
Answer:

Step-by-step explanation:
Connect points I and K, K and M, M and I.
1. Find the area of triangles IJK, KLM and MNI:

2. Note that

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.
Answer:
7 square miles
Step-by-step explanation:
I just counted the squares lol