If you use the first answer, the equation becomes 2x + 2y = 16. You could subtract this from equation Q to eliminate x.
The second answer makes equation P -2x - 2y = -16. You could add this to equation Q to eliminate x. You can use both of these methods to eliminate the x term, but if you can only choose one, you should choose the second option.
Answer:
A) This is correct. The angles of <em>AEB </em>are the angles of <em>DEC. </em>This means that this is true.
B) This is incorrect. This triangle is NOT an isosceles triangle. The angles of an isosceles triangle are supposed to be about 36° for the top angle, and 72° for the other two angles. (At least this is what I was told)
Hope this helps :)
-wait. .___. they both might be true.
According to G oogle: "All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90°. The sum of three angles of an isosceles triangle is always 180°, which means we can find out the third angle of a triangle if the two angles of an isosceles triangle are known."
Answer:
b
Step-by-step explanation:
Answer:
C = 5.
Step-by-step explanation:
First, you need to remember that:
For the function:
h(x) = Sinh(k*x)
We have:
h'(x) = k*Cosh(k*x)
and for the Cosh function:
g(x) = Cosh(k*x)
g'(x) = k*Cosh(k*x).
Now let's go to our problem:
We have f(x) = A*cosh(C*x) + B*Sinh(C*x)
We want to find the value of C such that:
f''(x) = 25*f(x)
So let's derive f(x):
f'(x) = A*C*Sinh(C*x) + B*C*Cosh(C*x)
and again:
f''(x) = A*C*C*Cosh(C*x) + B*C*C*Sinh(C*x)
f''(x) = C^2*(A*cosh(C*x) + B*Sinh(C*x)) = C^2*f(x)
And we wanted to get:
f''(x) = 25*f(x) = C^2*f(x)
then:
25 = C^2
√25 = C
And because we know that C > 0, we take the positive solution of the square root, then:
C = 5
