Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
Answer:
Step-by-step explanation:
Use the rise over run formula
y2-y1/x2-x1
1992-1990/3317-3038
2/-279
-139.5
Answer:
Figure (i) and (iv)
Step-by-step explanation:
Given:
Optional figure is given in attached file.
We need to find two figures that are similar to the 5 by 10 figure.
All the given figure are
form.
Where m represent the number of rows and n represent the number of columns.
Solution:
Observe that in the given figure 5 by 10, the number of rows is 5 and number of columns is 10, that is, the number of columns is double of that the number of rows.
So we need to find two such figures whose number of columns is double of the number of rows.
From the given figures, figure (i) the number of rows is 2 and number of columns is 4, which is double of number of rows. so it is similar to 5 by 10 figure.
Similarly in figure (iv), the number of rows is 4 and number of columns is 8. so the number of columns is double the number of rows, so it is similar to the figure 5 by 10.
Therefore, the two figures that are similar to 5 by 10 figure are given in attached file such as (i) and (iv).
√(43) = 6.55.
-3√(2) = -4.24...
√(43) > 3.6 > -3√(2)