
Here, we want to find the diagonal of the given solid
To do this, we need the appropriate triangle
Firstly, we need the diagonal of the base
To get this, we use Pythagoras' theorem for the base
The other measures are 6 mm and 8 mm
According ro Pythagoras' ; the square of the hypotenuse equals the sum of the squares of the two other sides
Let us have the diagonal as l
Mathematically;
![\begin{gathered} l^2=6^2+8^2 \\ l^2\text{ = 36 + 64} \\ l^2\text{ =100} \\ l\text{ = }\sqrt[]{100} \\ l\text{ = 10 mm} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20l%5E2%3D6%5E2%2B8%5E2%20%5C%5C%20l%5E2%5Ctext%7B%20%3D%2036%20%2B%2064%7D%20%5C%5C%20l%5E2%5Ctext%7B%20%3D100%7D%20%5C%5C%20l%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B100%7D%20%5C%5C%20l%5Ctext%7B%20%3D%2010%20mm%7D%20%5Cend%7Bgathered%7D)
Now, to get the diagonal, we use the triangle with height 5 mm and the base being the hypotenuse we calculated above
Thus, we calculate this using the Pytthagoras' theorem as follows;
use this formula Sn=n/2[2a+(n-1)d]
here,d=a2-a1 and n= no. Of terms
Step-by-step explanation:
S8=8/2[2(1)+(8-1)2]
S8=64
Step-by-step explanation:
Part 1:
Rotation counterclockwise 90 °ree; of a point is
Here is a table of the transformation of all four points:
P(-7,0)-> P'(-0,-7) =P'(0,-7)
A(-4,2)-> A'(-2,-4)
R(-3,0)-> R'(0,-3)
K(-6,2)-> K'(-2,-6)
The initial and rotated images are shown attached.
Part 2: it needs to be done with the technology tool that you have. We can check your work if you post what you did.
Answer:
-156?
Step-by-step explanation: