Answer:
363.22
Step-by-step explanation:
<u>Method 1: </u>
You could find the whole figure surface area than divided by 1/2
<u>Method 2:</u> (the one I'm going to personally be doing)
Break the figure into two rectangular figures
Formula for surface area of rectangular prism:
A = 2(width x length + height x length + height x width)
Figure 1:
A = 2(width x length + height x length + height x width)
height = 3.8 yd
length = 10.1 yd
width = 4.3 yd
A = 2((4.3) x (10.1) + (3.8) x (10.1) + (3.8) x (4.3))
A = 2(98.15)
A = 196.3
Figure 2:
A = 2(width x length + height x length + height x width)
height = 8.4 yd
length = 10.1 yd
width = 2 yd
A = 2((2) x (10.1) + (8.4) x (10.1) + (8.4) x (2))
A = 2(121.84)
A = 243.68
There is overlapping surface area that shouldnt be include so we need to subtract it...
<u>For one face of figure 1</u>
3.8 x 10.1 = 38.38
Total:
Figure 1 + Figure 2 - 2(one face)
196.3 - 38.38 = 157.92
243.68 - 38.38 = 205.3
205.3 + 157.92 = 363.22
Answer: 0 and 1, in that order
The numbers <u> 0 </u> and <u> 1 </u> are respectively the additive and multiplicative identities of rational numbers.
===========================================================
Explanation:
The additive identity is 0 because adding 0 to any number leads to the original number. For instance, 7+0 = 7. In general we can say x+0 = x or we could also say 0+x = x.
The multiplicative identity is 1 because multiplying 1 with anything leads to that original number. Example: 1*5 = 5 or 9*1 = 1. The general template is x*1 = x which is the same as saying 1*x = x.
These ideas not only apply to rational numbers, but to real numbers as well.
7 7/9 clergs.
wergs= w
nergs= n
clergs= c
3w= 4n, so 5w= 6 2/3n
Since 6n=7c, then 6 2/3n= 7 7/9n
Hope that answered your question!!
If a 15 lbs dog eats................................17/8 cups of food
a 34 lbs dog will eat,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,? cups
(34 * 17/8)/15=(34*17)/(15*8)=289/15= 19 4/15 cups of food
Answer:
No invariant point
Step-by-step explanation:
Hello!
When we translate a form, in this case a polygon We must observe the direction of the vector. Since our vector is:
1) Let's apply that translation to this polygon, a square. Check it below:
2) The invariant points are the points that didn't change after the transformation, simply put the points that haven't changed.
Examining the graph, we can see that no, there is not an invariant point, after the translation. There is no common point that belongs to OABC and O'A'B'C' simultaneously. All points moved.
