How do i total this please help 8.3m 5.2m 4.36m 42.1m 86.1m

An animal shelter houses dogs, cats, and rabbits. there are 126 animals at the shelter. of the animals, 1/3 are cats. three four

viva [34]

Given that an animal shelter houses three kinds of animals: dogs, cats and rabbits and it has a total of 126 animals. 1/3 of the animals are cats. This means, 1/3 of 126 is 42. So there are 42 cats. 3/4 of the remaining animals, which is 84 are dogs. So 3/4 of 84 is 63. So there are 63 dogs. 63 dogs plus 42 cats is 105 animals. 126 minus 105 is 21. Therefore, there are only 21 rabbits. Hope this helps.

6 0

3 years ago

Determine the length of the longest metal rod which could be stored in a rectangle box 20 cm by 50 cm by 30 cm.

navik [9.2K]

Answer:

61.64 cm

Step-by-step explanation:

The longest metal rod which could be stored in a rectangular box will be equal to its diagonal:

$\therefore \: length \: of \:metal \: rod \\ = \sqrt{ {l}^{2} + {b}^{2} + {h}^{2} } \\ = \sqrt{ {20}^{2} + {50}^{2} + {30}^{2} } \\ = \sqrt{400 + 2500 + 900} \\ = \sqrt{3800} \\ = 61.64414 \\ = 61.64 \: cm$

5 0

3 years ago

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Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clu

dmitriy555 [2]

Answer:

The figure is attached and the total distance is 1031 feet.

Step-by-step explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

$d_{T}=[d_{FP\ to\ Clue1}+d_{Clue1\ to\ Clue2}+d_{Clue2\ to\ Clue3}]*distance\ per\ unit\\d_{T}=[\sqrt{(Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2}+\sqrt{(Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2}+\sqrt{(Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2}]**distance\ per\ unit\\$ Substituting the values

$d_{T}=[\sqrt{(0-0)^2+(5-0)^2}+\sqrt{(6-0)^2+(0-5)^2}+\sqrt{(0-6)^2+(-5-0)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{(0)^2+(5)^2}+\sqrt{(6)^2+(-5)^2}+\sqrt{(-6)^2+(-5)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{0+25}+\sqrt{36+25}+\sqrt{36+25}]*50 \text{ feet}\\d_{T}=[\sqrt{25}+\sqrt{61}+\sqrt{61}]*50 \text{ feet}\\d_{T}=[5+7.81+7.81]*50 \text{ feet}\\d_{T}=[20.62]*50 \text{ feet}\\d_{T}=1031 \text{ feet}\\$