A. Which reduction should she use so the picture fills as much of the frame as possible, without being too large?
Find the scale factor to get rom 7 1/3 inches to 5 1/3 inches:
5 1/3 / 7 1/3 = 0.7272
Now rewrite the fraction as decimals:
2/3 = 0.667
¾ = 0.75
5/9 = 0.555
The closest scale that would still fit the frame would be 2/3 because it is under 0.727.
B. How much extra space is there in the frame when she uses the reduction from Part A?
Multiply the original size by the scale factor to use:
7 1/3 x 2/3 = 4 8/9
Now subtract the scaled size from the original size:
7 1/3 – 4 8/9 = 2 4/9 inches extra
C. If she had a machine that could reduce by any amount, so that she could make the reduced picture fit in the frame exactly, what fraction would the reduction be?
Convert the scale from part A to a fraction:
0.72 = 72/99 which reduces to 8/11
1000 meters= 1 kilometer so your answer is 1.5 kilometers
{x,y}={24,12} I think that is the answer to this question....
The answer to this question is Letter B.
Answer:
<u>Option C</u>
Step-by-step explanation:
Given 0.5≤ x ≤ 0.7
According to the range of x, we will check which option is true
Let x = 0.6
A. 
If x = 0.6
∴0.6/3 > √0.6 ⇒ 0.2 > 0.77 ⇒ <u>Wrong inequality </u>
B. x³ > 1/x
If x = 0.6
∴ 0.6³ > 1/0.6 ⇒ 0.216 > 1.667 ⇒ <u>Wrong inequality </u>
<u></u>
C. √x > x³
If x = 0.6
∴ √0.6 > 0.6³ ⇒ 0.77 > 0.216 ⇒ <u>True inequality </u>
D. (x/3) > (1/x)
If x = 0.6
∴ 0.6/3 > 1/0.6 ⇒ 0.2 > 1.667 ⇒<u> Wrong inequality </u>
<u>So, The true inequality is option C</u>
