I am pretty sure you can only make one.
If two sides are known, and one of the angles, then the other bits can be deduced and are fixed.
Answer:
(4x)^ 10
Step-by-step explanation:
Question 1:
To start off this question, we can tell that this is a square because it has 4 right angles and 4 congruent sides.
A square has four parallel sides and 4 congruent sides, so a square is a rhombus and parallelogram.
A square has 4 right angles, so it's also a rectangle.
A square has 4 sides, so it's also a quadrilateral.
The first choice is your answer.
Question 2:
Not all quadrilaterals are rectangles, so A is incorrect.
Not all quadrilaterals are squares, so B is incorrect.
All rectangles are types of quadrilaterals, so C is correct.
Not all quadrilaterals are parallelograms, so D is incorrect.
Thus, C is your answer.
Question 3:
The first choice will not work because a rhombus will satisfy those conditions, and a rhombus is not always a square.
The second choice will work because only a square will satisfy that condition because only squares have 4 congruent sides along with equal diagonals.
Thus, the second choice is your answer.
Have an awesome day! :)
13 you add both of the factors
Answer: 7. g = -50, 8. f= 340/6 9. 3 10 .125
Step-by-step explanation:#7 g(3): 4(3)^2 +6 =12^2 +6 = 1444=6= 150. We divide 150 by 3 giving us -50. #8We start by putting f(6): -3(6)^2 -4(6) +8. We use PEMDAS. -18^2 -24 +8 = _324-24+8 F= 340/6 #9 g(15) = Square root of 15-6=9 and the square root of 9 is 3, therefore, the answer is 3. #10 h (x) = 2^x we plug -3 to 2^-3. It is a negative number and it gives us .125 or 12.5
