She can have 11 rows with 3 in the last row and she can have 6 rows with 3 in the last row
Given:
A rectangle has a length 5 meters more than five times the width.
The area of the rectangle is less than 100 meters squared.
To find:
The expression or inequality that represents all possible widths of the rectangle.
Solution:
Let x be the width of the rectangle.
Length of the rectangle is 5 meters more than five times the width.
Area of rectangle is
The area of the rectangle is less than 100 meters squared.
Divide both sides by 5.
It is true if one factor is negative and other is positive. So,
...(i)
...(ii)
Using (i) and (ii), we get
Therefore, the required expression or inequality for possible
widths of the rectangle is .
Answer: 2 , 1/3 cups
Step-by-step explanation:
1/3 is the base number and 2/3 is another third being added onto the singular 1/3. The awnser would be 2 because 1/3+1/3= 2/3 or 2/3/2= 1/3
(-3, -3) and (2, 7).
The solutions to a graphed system of equations are where the two graphs cross each other. So in this case the two equations cross at (-3, -3) and (2, 7), and those points are the solutions to the system of equations.
Answer:
90 clockwise (or counterclockwise) rotation and then a reflection over the axis between the two shape (those two steps go in any order)
Step-by-step explanation:
for this lets mark the innermost point of each shape a (blue or A) and a' (red or B)* and the second point b and b'
here we see that the two shapes are in a position to where they seem reflected over a non-existent third diagonal axis, though this is not the case, we need to bring the shape into a position where it can be transformed to the quadrant of shape B and overlap the shape
so when you have a reflection over a diagonal axis, we can rotate or reflect the shape to a new quadrant, and perform the step thats not the first, so say we made a reflection over the X-axis, the shape is now in the lower half of the graph with shape B, from here we perform our last step wich is to rotate the shape into the quadrant of shape B in a clockwise motion, now a and a' overlap and b and b' overlap, same for c, c',d and d'
(*the ' in this case is called a prime symbol, when used, distinguishes two points or lines on a graph, A' = A prime)
