Answer:
5 folders and 5 pens = yes
6 pens and 6 erasers = yes
1 pen and 4 notebooks = no
3 folders and 7 erasers = no
4 folders and 2 notebooks = yes
Step-by-step explanation:
Max total = $10
(5 folders x 1.29) + (5 pens x 0.70)
6.45 + 3.50 = 9.95
(6 pens x 0.70) + (6 erasers x 0.89)
4.20 + 5.34 = 9.54
(1 pen × 0.70) + (4 notebooks × 2.35)
0.70 + 9.40 = 10.10
(3 folders × 1.29) + (7 erasers × 0.89)
3.98 + 6.23 = 10.21
(4 folders × 1.29) + (2 notebooks × 2.35)
5.16 + 4.70 = 9.86
Answer:
Answer
2x+3y=14,2x−3y=2
Adding :
2x+3y=14
2x−3y=2
__________
4x=16
x=4
∴xy=4×2
=8
2×4+3y=14
8+3y=14
3y=6
y=2
Step-by-step explanation:
here you go
So in your circle their is a terminal point who is assigned by the value of P(x,y) and lies or determined by teh value of T=4pi/3. To get the coordinates of this terminal point first is to know what is the angle of the T and the answer is (-1/2, -sqrt(3)/2)
Answer:
The figure is NOT unique.
Imagine the following quadrilaterals:
Rectangle
Square
We know that:
Both quadrilaterals have at least two right angles.
However, they are not unique because they depend on the lengths of their sides.
Step-by-step explanation:
To construct a quadrilateral uniquely, five measurements are required. A quadrilateral can be constructed uniquely if the lengths of its four sides and a diagonal are given or if the lengths of its three sides and two diagonals are given.
Just given two angles we cannot construct a unique quadrilateral. There may be an infinite number of quadrilaterals having atleast two right angles
Examples:
All squares with varying sides
All trapezoids with two right angles
All rectangles with different dimensions
and so on.
Answer is
No.
