If the original side length is "s" and the original slant height is "h", the original surface area is
.. S = (base area) +(lateral area)
.. S = s² +(1/2)*(4s)*h
.. S = s(s +2h)
Now, if we make these replacements: s ⇒ 3s, h ⇒ h/5, we have
.. S' = (3s)(3s +2h/5)
.. S' = 9s² +(6/5)s*h . . . . . . . the formula for the modified area (in terms of original dimensions)
_____
Of course, in terms of the modified dimensions, the formula is the same:
.. S' = s'(s' +2h')
Answer:

Step-by-step explanation:
You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.
You will find when you look on your trig identity sheet that

so we will make that replacement, getting everything in terms of sin:

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

We can factor out the sin(theta), since it's common in both terms:

Because of the Zero Product Property, either
or

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

The next equation needs to first be solved for sin(theta):
so
and

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

Step-by-step explanation:
if you calculate both of you should ask her
Answer:
sorry you cant get help but you can help me!
GO TO MY PAGE
Step-by-step explanation: