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Usimov [2.4K]
3 years ago
14

PLZ HURRY IT'S URGENT!!!

Mathematics
2 answers:
luda_lava [24]3 years ago
8 0
C

Explanation
35 divided by 35 is 1 so as 56 divided by 56 so it would be 1:1
vladimir1956 [14]3 years ago
5 0

Answer:

Step-by-step explanation:

5/8

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Felipe picked a pumpkin that weighed three times the weight of Meg's pumpkin Meg's pumpkin with half the weight of Ryan's pump R
Xelga [282]

Answer:

21 lb

Step-by-step explanation:

Divide the amount of Ryan's pumpkin by 2 then multiply it by 3. Ryan's pumpkin weighs 14 lb divided by 2 is 7 then 7 multiplied by 3 is 21.

6 0
3 years ago
If the square root of 11025 is 105, then find the Square root of 1.1025?​
german

Answer:

→ √1.1025 is 1.05.

Step-by-step explanation:

Given that

√11025 = 105

Now,

√1.1025

⇛ √(11025/10000)

⇛ √(105²/100²)

⇛ √(105/100)²

⇛ 105/100

⇛ 1.05

∴ the square root√1.1025 is 1.05.

4 0
3 years ago
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A skydiver descended at a constant rate after opening her parachute. Three minutes after she opened her parachute, she had
Reptile [31]

Answer:

1056 feet per a minute

Step-by-step explanation:

math

3 0
3 years ago
Arnold needs to simplify the expression below.
Lunna [17]

Answer:

0.5x7

Step-by-step explanation:

Assuming that the original expression is 4.3*3+ 11(8.2-0.5*7)-6, according to PEMDAS multiplication takes priority over addition and subtraction,, and therefore, from the choices given, 0.5 * 7 must be performed first.

3 0
3 years ago
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8. Let R be the relation on the set of all sets of real numbers such that SRT if and only if S and T have the same cardinality.
sergejj [24]

Answer:

We must prove that the relation is reflexive, symmetric and transitive. Recall that to sets have the same cardinality if there exist a bijective mapping between them.

<em>Reflexive: </em>Take the identity map I:S\rightarrow S, which is bijective. Then SRS.

<em>Symmetric:</em> If SRT then, there exist a bijective map f:S\rightarrow R. In order to prove that TRS just take the inverse map of f: f^{-1} which is also bijective. Therefore, TRS.

<em>Transitivity: </em>Suppose that SRT and TRU. Also, assume that f is the bijective map between S and T, and g the bijective map between T and U. It is not difficult to check that the map h=g(f) is bijective and h:S\rightarrow U. Therefore, SRU.

Hence, the relation R is an equivalence relation.

The equivalence class of the set {0,1,2} is the class of all the sets with three elements, and we can associate it with the number 3. There is a construction of natural numbers based on this idea.

The equivalence class of Z is the same equivalence class of N. Therefore, is the class of all denumerable or countable sets.

Step-by-step explanation:

When we want to prove that a given relation R is equivalence, we need to check that R satisfies all the three conditions: reflexive, symmetric and transitivity. Usually the first two are very simple to prove and comes directly from the definition. The transitivity is more tricky. In this case we need to recall the definition of cardinality.

7 0
3 years ago
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