b^2-4ac<0 b^2-144<0 (b-12)(b+12)<0
-12<b<12
Answer:
15-9z
Step-by-step explanation:
combine like terms,, the like terms we have are the terms (z) and the whole numbers so we combine 6,8, and 1-- and combine -5z and -4z to get the solution.
Using the circle theorems, we have proven that m ∠RTW = 15°
<h3>Circle theorems </h3>
From the question, we are to prove that m ∠RTW = 15°
In the given diagram,
measure of arc ST = 30°
∴ m ∠SRT = 30°
m ∠SRT = ∠T + ∠W ( <em>Exterior angle of a triangle equals the sum of the two remote angles</em>)
Also,
∠T = ∠W (<em>Radii of the same circle</em>)
∴ m ∠SRT = ∠T + ∠T
m ∠SRT = 2 × ∠T
30° = 2 × ∠T
∠T = 30° /2
∠T = 15°
∴ m ∠RTW = 15°
Hence, we have proven that m ∠RTW = 15°
Learn more on Circle theorems here: brainly.com/question/27111486
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Given: x^2 - 6x + 2
Statements:
1) The graph of the quadratic equation has a minimum value:TRUE. WHEN THE COEFFICIENT OF X^2 IS POSITIVE THE PARABOLA OPEN UPWARDS AND ITS VERTEX IS THE MINIMUM.
2) The extreme value is at the point (3 , - 7): TRUE
You have to find the vertex of the parabola:
x^2- 6x + 2 = (x - 3)^2 - 9 + 2 = (x - 3)^2 - 7 => vertex = (3, -7)
3) The extreme value is at the point (7, -3): FALSE. THE RIGHT VALUE WAS FOUND IN THE PREVIOUS POINT.
4) The solutions are x = - 3 +/- √7. FALSE.
Solve the equation:
(x - 3)^2 - 7 = 0 => (x - 3)^2 = 7 => (x - 3) = +/- √7 => x = 3 +/- √7
5) The solutions are x = 3 +/- √7. TRUE (SEE THE SOLUTION ABOVE).
6) The graph of the quadratic equation has a maximum value: FALSE (SEE THE FIRST STATEMENT).
