The function f is continuous where f(−5) = −1 and f(5) = 6, and g is a function defined by g(x) = 1 − (f(x))2. Is there a value c for −5 < c < 5 such that g(c) = 1? Why, or why not? Yes; the function g is continuous
Yes; g(−5) < 1 < g(5), so IVT guarantees there is a value c for −5 < c < 5 such that g(c) = 1
No; 1 is not between g(−5) and g(5), so IVT cannot guarantee there is a value c for −5 < c < 5 such that g(c) = 1
No; the function g is not continuous

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180

6(h+1)/2=a

a=area

step-by-step explanation:

its the same thing as a square or rectangle but after multiplying two sides you need to divide it by two.

this is a good question! i’ll do some research and what i find, i’ll contact you asap when i have something! ✊