Answer:
Step-by-step explanation:
13). Area of a square = (Side)²
= (BC)²
Since, diagonals of a square bisect each other at 90°,
ΔBOC is a right triangle.
By applying Pythagoras theorem in the given triangle,
BC² = OB² + OC²
BC² = 2(OB)²
BC² = 2(7√2)²
BC = 
Area of square ABCD = (BC)²
= (√196)²
= 196 units²
14). Measure of interior angles of the regular hexagon = 120°
Area of the regular hexagon = 
From the given picture,
m∠BAC = m∠ABC = m∠ACB = 60°
Therefore, ΔABC is an isosceles triangle.
And all sides of this triangle will be equal in measure.
AB = AC = BC = 9 units
Area of the given regular hexagon = 
= 210.44 square units
≈ 210.4 square units
Answer:
x = 30, x = 5 and x = 9
Step-by-step explanation:
The diagonals of a square are perpendicular bisectors of each other, so
∠ AEB = 90° , then
3x = 90 ( divide both sides by 3 )
x = 30
---------------------------------------------------------
(9)
The 4 angles of a square are right and the diagonals bisect the angles, then
∠ BAC = 45° , so
9x = 45 ( divide both sides by 9 )
x = 5
-------------------------------------------------------
(10)
All sides are congruent , so
CD = AB , that is
3x - 5 = 2x + 4 ( subtract 2x from both sides )
x - 5 = 4 ( add 5 to both sides )
x = 9
Answer:
B & D
Step-by-step explanation:
The quadratic formula can be used to solve for x for any quadratic. Recall a quadratic is any function whose highest exponent known as degree is 2. Looking at the answer selections, A has one term with exponent 3. This isn't possible in the formula. Looking at answer C, the term with exponent 2 has the same coefficient on each side of the equal sign. When it is rearranged these will cancel out no exponent of 2 will be in the equation anymore. Only B & D work after being rearranged and simplified
Answer:
Graphs 1, 2, 3
Step-by-step explanation:
Graph 1 represents a <u>linear function</u> with a domain containing all real numbers, as any input substituted into the equation or function will produce a corresponding output. The arrows on the opposite ends of the line represents the infinite input values that have its corresponding output values.
Graph 2 represents a <u>quadratic function</u> with a domain containing all real numbers, as it does not have any constraints in terms of input values. As it opens up, the graph of the parabola infinitely widens (horizontally).
Graph 3 represents an <u>absolute value function</u> with a domain containing all real numbers. Similar to the explanation for graph 2 on quadratic functions, the downward-facing graph of the given absolute value function widens infinitely horizontally, as it does not have any constraints on input values.
