True!
all parabolas with vertical directrix DO open to the left/right.
Answer:
30
Step-by-step explanation:
200x.15=30
200-30=170
Answer:
Step-by-step explanation:
30. Given: rectangles QRST and RKST
Prove: ΔQSK is isosceles
An isosceles triangle is a triangle which has two sides and two angles to be equal.
Thus,
From rectangle QRST, the diagonals of rectangles are similar.
i.e RT ≅ QS (diagonal property)
Also, RT ≅ SK (opposite sides of rectangle RKST)
Thus,
RT ≅ QS ≅ SK
Therefore,
ΔQSK is an isosceles triangle.
31. Given: Rectangles QRST, RKST and JQST
Prove: JT ≅ KS
From rectangle QRST, the diagonals of rectangles are similar.
i.e RT ≅ QS (diagonal property)
But,
JT // QS and RT // KS
Thus,
JT ≅ QS (opposite sides of rectangle JQST)
also,
RT ≅ KS (opposite sides of rectangle RKST)
So that,
JT ≅ QS ≅ RT ≅ KS
Therefore,
JT ≅ KS
You did not provide any question. I will give general advice.
Pythagorean Theorem: a² + b² = c²
If you know two sides, you know ll three sides.
A house sits in the shade.
The house is 18 feet tall. The shadow is 24 feet long. What is the hypotenuse of this triangle?
We know a and b here! a=18, b=24.
18 * 18 = 324.
24 * 24 = 576.
324+576 = 900
The square root of 900 is what will be your answer:
18*18 + 24*24 = 30*30.
I hope this helps!
Answer:
Option B. 5.5 ft²
Step-by-step explanation:
Perimeter of the square piece of land = length of fence required
4(length of a side of the square) = 24 ft
Length of a side = 6 ft
Area of the square land = (Side)²
= 6²
= 36 square ft
Area of the side walk = Area of the square - [Area of two triangles separated by the sidewalk]
= 36 - [1/2 (Area Of The Sqaure) + 1/2 (Base) (Height) ]
= 36 - [36/2 + 1/2 (6 - 1) (6 - 1)]
= 36 - (18 + 12.5)
= 36 - 30.5
= 5.5 ft²
Therefore, Option B. is the correct answer.
