Question

1. communicate and justify suppose you plot the locations of the animals on a number line. which animal would be represented by the point farthest from 0 on the number line? explain.
2. which animal is closest to a depth of -\frac{7}{10} km?
3. the change in the value of a stock is represented by the rational number - 5.90. describe, in words, what this means.
4. check for reasonableness a classmate ordered these numbers from greatest to least. is the classmate correct? construct an argument to justify your answer. 4.4, 4.2, -4.42, -4.24
5. analyze and persevere a textile artist is asked to make a wall - hanging that measures at most 3.75 feet wide by 6.5 feet tall. when finished, the wall hanging measures 3\frac{3}{8} feet wide by 6\frac{3}{4} feet tall. does the wall hanging meet the specifications? explain.
6. higher order thinking suppose \frac{a}{b}, \frac{c}{d}, and \frac{e}{f} represent three rational numbers. if \frac{a}{b} is less than \frac{c}{d}, and \frac{c}{d} is less than \frac{e}{f}, compare \frac{a}{b} and \frac{e}{f}. explain.

animal | possible locations relative to oceans surface
bloodbelly comb jelly | -\frac{8}{10} km
deep sea anglerfish | -\frac{2}{3} km
fanfin anglerfish | -2\frac{1}{4} km
gulper eel | -1\frac{1}{10} km
pacific blackdragon | -\frac{3}{10} km
slender snipe eel | -\frac{6}{10} km

Explanation:

25.

Step1: Find absolute - values

The absolute - value of a number represents its distance from 0 on the number line. Calculate the absolute - value of each depth.
$|-\frac{8}{10}| = 0.8$, $|-\frac{2}{3}|\approx0.67$, $|-2\frac{1}{4}| = 2.25$, $|-1\frac{1}{10}| = 1.1$, $|-\frac{3}{10}| = 0.3$, $|-\frac{6}{10}| = 0.6$

Step2: Compare absolute - values

Compare the absolute - values: $0.3<0.6<0.67<0.8<1.1<2.25$

Step1: Calculate differences

Find the absolute difference between $-\frac{7}{10}=-0.7$ and each of the given depths.
For bloodbelly comb jelly: $|-\frac{7}{10}-(-\frac{8}{10})|=|-\frac{7}{10}+\frac{8}{10}|=\frac{1}{10} = 0.1$
For deep - sea anglerfish: $|-\frac{7}{10}-(-\frac{2}{3})|=|-\frac{7}{10}+\frac{2}{3}|=|-\frac{21}{30}+\frac{20}{30}|=\frac{1}{30}\approx0.033$
For fanfin anglerfish: $|-\frac{7}{10}-(-2\frac{1}{4})|=|-\frac{7}{10}+\frac{9}{4}|=|-\frac{14}{20}+\frac{45}{20}|=\frac{31}{20}=1.55$
For gulper eel: $|-\frac{7}{10}-(-1\frac{1}{10})|=|-\frac{7}{10}+\frac{11}{10}|=\frac{4}{10} = 0.4$
For Pacific blackdragon: $|-\frac{7}{10}-(-\frac{3}{10})|=|-\frac{7}{10}+\frac{3}{10}|=\frac{4}{10} = 0.4$
For slender snipe eel: $|-\frac{7}{10}-(-\frac{6}{10})|=\frac{1}{10}=0.1$

Step2: Compare differences

Compare the differences: $0.033<0.1 = 0.1<0.4 = 0.4<1.55$

The negative sign in front of the rational number $- 5.90$ indicates a decrease.

Answer:

The fanfin anglerfish, with a depth of $-2\frac{1}{4}$ km, is farthest from 0 on the number line because its absolute - value $|-2\frac{1}{4}| = 2.25$ is the largest among the given values.

26.