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White raven [17]
3 years ago
9

In a right triangle ABC with the right angle at C, TanA=0.75 What is TanB?

Mathematics
1 answer:
n200080 [17]3 years ago
4 0
Tan B=1.33. If you put tan^-1 and find the angle of A, you’ll get 36.87, so 90-36.87=53.13, then put Tan53.13, you’ll get TanB=1.33
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400:500 in simplest form
Inessa05 [86]

Answer:

4/5

Step-by-step explanation:

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3 years ago
Mary and Roberto go to the movies. They buy a large popcorn for $8.50, and two medium sodas for $4.25. How much did they spend a
alukav5142 [94]

Answer:

$19.15

Step-by-step explanation:

Equation 1: 8.50+4.25+4.25= 17.00  (price of the popcorn and drinks)

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Answer the following true or false. Justify your answer.
Anna007 [38]

Answer:

a) False b) False c) True d) True e) True

Step-by-step explanation:

a) If A is a subset of B, and x∈B, then x∈A. False

Suppose A is Z (Set of Integers), and B is R (Set of Real Numbers). Then A is a subset of B. x ∈ B, let's say x equals π. If x ∈ B (B = Real Numbers) and x=π then x ∉ A (Z).

We could call A, any other subset of Real numbers (Q, I,..) and we both would come up to the same conclusion when it comes to Real numbers the Set.

So this conclusion is False. Not always an element of a subset is an element of a set.

b) False

For this one, I've drawn some lines, and it is useful to work with them.

Given the set {(x,y) ∈ R²| 0<x<0} then all negative and positive numbers but zero belong to this set.

c) If A and B are square matrices, then AB is also square. True

Taking into account the rules for multiplying matrices. The number of columns of A must be the same number of lines of B to there can be a matrix product.

Whenever we multiply square matrices, we'll always get square matrices. Then this conclusion is true.

d) A and B are subsets of a set S, then A∩B and A∪B are also subsets of S. True

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A∩B = Z∩Q=∅ and A∪B =Z∪Q = subset

Since the ∅ empty set ⊂ in every set and ZUQ is another Subset of R this is a True conclusion.

e) True. For a matrix A, we define A²= AA

For any Power of Matrices, all we have to do is multiply any given matrix by itself for a given number of times.

M²=M*M

M³=M*M*M

7 0
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