PLEASE HELP
Determine a quadratic function f(x) = ax2 + bx + c such that the maximum value of the function is 2 and the graph of the function has x-intercepts of 3 and 5.
A) f(x) = x2 − 8x + 15
B) f(x) = 2(x2 − 8x + 15)
C) f(x) = −2(x2 − 8x + 15)
D) f(x) = −2(x2 − 8x − 15)

c

step-by-step explanation:

at least i think its

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation:

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3