Answer:
The correct locations are;
Part to part ratio is 3 cats to 5 dogs
Whole to part ratio is 10 pets to 2 rabbits
Part to whole ratio is 5 dogs to 10 pets
Step-by-step explanation:
The given information are;
The number of cats Jim has = 3
The number of dogs Jim has = 5
The number of rabbit Jim has = 2
The total number of pets Jim has = 3 + 5 + 2 = 10 pets
Therefore;
The fraction of the pets that are cats = 3/10
The fraction of the pets that are dogs = 5/10
The fraction of the pets that are rabbits= 2/10
Therefore we have;
Part to part ratio = 3/10 cats to 5/10 dogs = 3 cats to 5 dogs
Whole to part ratio → 10 pets to 2 rabbits
Part to whole ratio → 5 dogs to 10 pets.
Answer:
32 miles (I'm sure you can convert to 32 without units)
Step-by-step explanation:
We can simply add up all of the values, yielding 25 + 6 + 1 = 32 total miles.
Answer:
x = 5/4
Step-by-step explanation:
32^(x-1) = 16^(x/4)
(2^5)^(x-1) = (2^4)^(x/4) . . . . . express as powers of 2
2^(5x -5) = 2^x . . . . . . simplify
5x -5 = x . . . . . . equate exponents
4x = 5 . . . . . . . . add 5-x
x = 5/4
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Applicable rules of exponents are ...
(a^b)^c = a^(bc)
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Since this solution does not match any offered answer choice, we suggest you ask your teacher to show you how to work this problem.
Answer:
115°
Step-by-step explanation:
● DAG is equal to 180°
● DAG = DAH + HAG
● HAG IS 130°
● DAG = DAH + 130° => DAH + 130° = 180°
● DAH = 180° - 130° = 50 °
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ADH is a triangle
● ADH + DHA + HAD = 180°
● ADH + DHA + 50° = 180°
DHA shares the same vertex with CHE. They are opposite angles so they have the same size (35°)
● ADH + 35° + 50° = 180°
● ADH + 85° = 180°
● ADH = 180° - 85°
● ADH = 95°
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BDA equals 180°
● BDA = 180°
● BDH + ADH = 180°
● BDH + 95° = 180°
● BDH = 180° -95°
● BDH = 85°
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BDE is a traingle
● BDH+ DEB + EBD = 180°
● 85° + DEB + 30° = 180°
● DEB + 115° = 180°
● DEB = 180° - 115°
● DEB = 65°
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CEF equals 180°
● CEF = 180°
● DEB + x° = 180°
● 65° + x° = 180°
● x° = 180° - 65°
● x° = 115°
Answer:
D) (4,5)
Step-by-step explanation:
the ordered pair (4, 5) is the only one that, when substituted into each inequality statement, made both of them true