Answer:
See verification below
Step-by-step explanation:
Remember that the Intermediate Value Theorem (IVT) states that if f is a continuous function on [a,b] and f(a)<M<f(b), there exists some c∈[a,b] such that f(c)=M.
Now, to apply the theorem, we have that f(0)=0²+0-1=-1, f(5)=5²+5-1=29, then f(0)=-1<11<29=f(5). Additionally, f is continous since it is a polynomial. Then the IVT applies, and such c exists.
To find, c, solve the quadratic equation f(c)=11. This equation is c²+c-1=11. Rearranging, c²+c-12=0. Factor the expression to get (c+4)(c-3)=0. Then c=-4 or c=3. -4 is not in the interval, then we take c=3. Indeed, f(3)=3²+3-1=9+3-1=11.
Answer:
f(2) = 19
Step-by-step explanation:
2(3²) + 1 = 2(9) + 1 = 19
It depends on the type of triangle like say it was a right triangle then a+b=c ok so you a=11 your b=18 then all you have to do is find you what c=s so c=x i hope this helped
By Angle-Angle simlilarity postulate :
If two angles of one triangle congruent to two angles of another, then triangles must be similar.
So, I think the answer is
<span>All isosceles triangles are not similar. The pair of congruent angles within one triangle is not necessarily congruent to the pair of congruent angles within the other triangle.
Because two base angles in </span>isosceles triangle are congruent, but it could be a lot of isosceles triangles that have different congruent base angles.
For example,
45-45-90 is an isosceles triangle, and 30-30-120 is an isosceles triangles, but they do not have 2 congruent angles.
