Answer:
D) x = 4, y = -2, z = 3
Step-by-step explanation:
x = 3z − 5
2x + 2z = y + 16
2(3z - 5) + 2z = y + 16
6z - 10 + 2z = y + 16
8z = y + 26 ---> (A)
7x − 5z = 3y + 19
7(3z - 5) - 5z = 3y + 19
21z - 35 - 5z = 3y + 19
16z = 3y + 54 ---> (B)
8z = y + 26
16z = 3y + 54
2(y + 26) = 3y + 54
2y + 52 = 3y + 54
y = -2
8z = -2 + 26
8z = 24
z = 3
x = 3(3) - 5
x = 4
Answer:
75%
Step-by-step explanation:
20-5=15 - change in numbers
15/20×100%= 75%

- Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>
- To find - <u>Area </u><u>of </u><u>trapezium</u>
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
<u>Construction</u><u> </u><u>-</u>

Now , we can clearly see that AECD is a parallelogram !
AE = CD = 13 cm
Now ,

Now , In ∆ BCE ,

Now , by Heron's formula

Also ,

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

hope helpful :D