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denpristay [2]
3 years ago
5

3^4 - ( 9 + 6) * 5 + 4^0

Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

Answer:

43 is correct answer

Step-by-step explanation:

{3}^{4}  - (9 + 6) \times 5 +  {4}^{0}  \\  =  {3}^{4}  - 9 - 6 \times 5 + 1 \\  =   {3}^{4}  - 9 - 30 + 1 \\  =  {3}^{4}  - 39 + 1 \\  =  {3}^{4}  - 38 \\  = 81 - 38 \\  = 43

Note:

1. When ever anything ⁰ =1

Eg:4⁰ =1, 5⁰=1

2. Always follow bodmas rule

hope it helped you:)

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x+7+14x

x+14x+7

15x+7

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Given that the measure of ∠x is 149°, and the measure of ∠y is 110°, find the measure of ∠z.
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5 0
3 years ago
Read 2 more answers
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:

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4 0
3 years ago
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