Answer:
f(x) = -2x² - 8x - 2
General Formulas and Concepts:
- Order of Operations: BPEMDAS
- Expand by FOIL (First Outside Inside Last)
- Standard Form: f(x) = ax² + bx + c
- Vertex Form: f(x) = a(bx + c)² + d
Step-by-step explanation:
<u>Step 1: Define function</u>
Vertex Form: f(x) = -2(x + 2)² + 6
<u>Step 2: Find Standard Form</u>
- Expand by FOILing: f(x) = -2(x² + 4x + 4) + 6
- Distribute -2: f(x) = -2x² - 8x - 8 + 6
- Combine like terms (constants): f(x) = -2x² - 8x - 2
Answer:
x = 59°
y = 67°
Step-by-step explanation:
x = y - 8
x + y + 54 = 180
(y - 8) + y = 180 - 54
2y - 8 = 126
2y = 134
y = 67°
x = 59°
Step-by-step explanation:
<h2>
Answer:</h2>
Figure B
<h2>
Step-by-step explanation:</h2>
The Pythagorean Theorem is 
, where c is the longest side of the triangle (the hypotenuse).
To find the side length of each square, you have to square root the area of each square. This means that Figure A has side lengths of 3, 6 and 8 units. Figure B has side lengths of 5, 12 and 13 units.
In Figure A, if the triangle is right-angled, the equation 
must be correct. 9 + 36 = 45. 45 is not equal to 64, so the triangle is not right-angled.
In Figure B, if the triangle is right-angled, the equation 
must be correct. 25 + 144 = 169. 169 is 13 squared, so the triangle is right-angled.
Alternatively, as you are already given the square values for each side length, there is no need to square root and square again. You can just test if the two smaller areas equal the larger area, but the explanation above uses a more detailed example of the Pythagorean Theorem.
Answer:
m∠FEH = 44°
m∠EHG = 64°
Step-by-step explanation:
1) The given information are;
The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°
Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°
Therefore, angle subtended by arc ∠HGF at the center = 88°
The angle subtended by arc ∠HGF at the circumference = m∠FEH
∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)
m∠FEH = 44°
2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°
The sum of angles in a quadrilateral = 360°
Therefore;
m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)
m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°
m∠EHG = 64°.
