-4m+15+7m<15
First we have to simplify -4m and 7m
3m+15<15 /-15 (subtract 15 both sides)
3m+15-15<15-15
3m<0
m<0 - its the result.
In order to prove this, we have to put the trapezoid to the coordinate system. In the attached photo you can see how it has to be put. The coordinates for the vertices of trapezoid written according to the midpoint principle. By using the distance between two points formula, we can find the coordinates for the vertices of the rhombus.
and
. The coordinates of D is
and
. The coordinates of E is
Since we have the reflection in this graph, the coordinates of F is
And the coordinates of G is (0,0).
Using the distance formula, we can find that
Since all the sides are equal this completes our proof. Additionally, we can find the distances of EG and DF in order to show that the diagonals of this rhombus are not equal. So that it is not a square, but rhombus.
Pemdas if you don't know what they are it looks like this.. (4+5)-5+4
similar to that.....
I am pretty sure you wouldn't add that in a agenda because of the different properties it has but there are plenty more ways i just haven't figured them out
Answer: 102
Step-by-step explanation:
6x6 to get the square
7x6 to get part of the other shape
21 -6-7 to get the triangle that we don't have a number
8x6 to get a square but then divide by 2 too get the triangle.
now add all together to get 102
Answer:
22
Step-by-step explanation:
Pretend the 10 values in the first sentence are a,b,c,d,e,f,g,h,i,j
Pretend the addition 5 values is k,l,m,n,o
So the mean of all the 15 data is (a+b+c+d+e+f+g+h+i+j+k+l+m+n+o)/15=20
So the sum of all 15 data is a+b+c+d+e+f+g+h+i+j+k+l+m+n+o=300 since 15(20)=300
Now let's look at the first 10: We have their mean so we can write:
(a+b+c+d+e+f+g+h+i+j)/10=19
so a+b+c+d+e+f+g+h+i+j=190 since 10(19)=190
So that means using our first sum equation and our equation sum equation we have
190+k+l+m+n+o=300
k+l+m+n+o=300-190
k+l+m+n+o= 110
So the average of those 5 numbers mentioned in your problem is 110/5=22