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leva [86]
3 years ago
6

If sinA=a_1/a+1 find value of tanA​

Mathematics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

a

Step-by-step explanation:

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7.
Morgarella [4.7K]

Answer:

7

Step-by-step explanation:

4 0
3 years ago
Use the distributive property -5 {x + 4} + -3x + -9x + -7
grin007 [14]

Answer:

-17x - 27

Step-by-step explanation:

Distributive property is the multiplication of the number outside the parenthesis and the ones inside.

-5(x + 4) + -3x + -9x + -7

-5x - 20 + -3x + -9x + -7

Combine like terms:

-5x - 20 + -3x + -9x + -7

-5x +-3x

 -8x + -9x

      -17x

-17x - 20 + -7

     -20 + -7

         -27

The answer is -17x - 27

Hope this helped.

8 0
3 years ago
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Algebra 2
olga55 [171]

Answer:

=5

Step-by-step explanation:

-7-6(-2)

-7+12

=5

sorry didn't read it right

4 0
4 years ago
Are u smarter than a first grader
densk [106]

Answer:

404.48

Step-by-step explanation:

area is length times width

8 0
3 years ago
Read 2 more answers
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