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Allushta [10]
3 years ago
10

Help!!!

Mathematics
1 answer:
stepladder [879]3 years ago
6 0

Step-by-step explanation:

No, a similar shape have proportional side length.

the triangles side length aren't proportional.

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Which box plot represents a set of data that has the least mean absolute deviation?
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Answer:

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Step-by-step explanation:

7 0
3 years ago
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Why are there no properties of subtraction or division?
tia_tia [17]
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4 0
3 years ago
Select either relation (if the set is a relation but not a function), function (if the set is both a relation and a function), o
galina1969 [7]

Answer:  The correct options are

(A) relation

(B) function.

Step-by-step explanation:  We are given to check whether the following set is a relation or a relation and a function or neither :

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U = {1, 2, 3, 4}  and  V = {2}.

Then, we see that the Cartesian Product of U and V is given by

U × V = {(1, 2) (2, 2) (3, 2) (4, 2)}.

Since a relation is a subset of Cartesian product, so the given set A is a relation.

Also, we know that a relation is a function if every first element is associated with one second element.

Since the relation A satisfies this condition, so the relation A is a function.

Thus, the given set A is both a relation and a function.

Option (A) and (B) are correct.

6 0
3 years ago
What is the value of 4 in the number 358.547? A) 0.0004 B) 0.004 C) 0.04 D) 4
LiRa [457]
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3 0
3 years ago
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Solve this query plzzz​
Olin [163]

Step-by-step explanation:

\frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   = \tan \:A \\  \\ LHS =  \frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   \\  \\  =  \frac{2\sin \:2A.\cos \:2A}{\cos \:2A}   \times  \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\  \\  = 2\sin \:2A   \times  \frac{1 - 2 { \cos}^{2}A  +  1}{1 - 2 { \cos}^{2}2A  + 1 } \\  \\  = 2\sin \:2A   \times  \frac{2- 2 { \cos}^{2}A  }{2 - 2 { \cos}^{2}2A   } \\  \\   = 2\sin \:2A   \times  \frac{2(1 - { \cos}^{2}A)  }{2 (1-  { \cos}^{2}2A)   } \\  \\   = 2\sin \:2A   \times  \frac{1 - { \cos}^{2}A}{1-  { \cos}^{2}2A   } \\  \\     = 2\sin \:2A   \times  \frac{ { \sin}^{2}A}{{ \sin}^{2}2A   } \\  \\    = 2  \times  \frac{ { \sin}^{2}A}{{ \sin}2A   } \\  \\   = 2  \times  \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \:   A } \\  \\   = \frac{ { \sin}A}{ \cos \:   A }  \\  \\  = tan \: A \\  \\  = RHS \\

7 0
4 years ago
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