Part A: Net A is correct
Net B is incorrect because de triangular sides do not close the opening left in both sides.
Part B: AB=3 in., BC=5in., CD=8.6in.
Part C: The surface area of the prism is the area of the the big rectangle in the net + the area of the 2 triangles
Area of the big rectangle
8.6• ( 3+4+5)= 103.2 in ^2
Area of the triangles
If we get the 2 trangles together along their longest side we get another rectangle
3•4 =12 in^2
Surface area of prism is 103.2+12=115.2 in^2
Answer:
3.00/8
Step-by-step explanation:
because u simplify 15/5 and u get 3
Answer:
8
Step-by-step explanation:
6 divided by.75 if u think I'm wrong check and correct me
First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:

To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:

We have our answer: approximately
0.14 cm or
1.4 mm per year
Answer:
2(3a-2b+c)=2(3a)-2(2b)+2(c)=6a-4b+2c
Hope this helped ur urgency