QUESTION 5
The side length of the first tra-pezoid and the second tra-pezoid are in the ratio .
To find the corresponding ratio of their area, we square each term in the ratio.
The ratio of the area of the first figure, to the ratio of the area of the second is
QUESTION 6
The side length of the first figure and the second figure are in the ratio .
To find the corresponding ratio of their area, we square each term in the ratio.
The ratio of the area of the first figure, to the ratio of the area of the second figure is
QUESTION 7
Let represent the height of this triangle.
This implies that;
The area of the triangle is
The base is 9 ft and the height is 3 ft.
We substitute these values to obtain;
to the nearest tenth.
QUESTION 8
The area of a parallelogram is
Let h represent the height of this parallelogram.
The area of the parallelogram is
to the nearest tenth.
Solution is lr =Al
=(5p-8)= (2p+22)
=5p-8 =2p+22
=5-2p= 22+8
=3p= 30
P=30/3= 10 so p =10
Answer:
x=-1, y=4. (-1, 4).
Step-by-step explanation:
-6x-y=2
-7x-4y=-9
----------------
y=-6x-2
-7x-4(-6x-2)=-9
-7x+24x+8=-9
17x=-9-8
17x=-17
x=-17/17
x=-1
-6(-1)-y=2
6-y=2
y=6-2=4
Rita spent 20$. By multiplying 5 pairs with 4$ we can see that it will equal 20$
I am pretty sure R is -19 meters and N is -7 and A is 476 AD