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faust18 [17]
3 years ago
13

7 + (2 + 6) 2 ÷ 4 ⋅ (1/2)^4 A: 23 B: 18 C: 9 D: 8

Mathematics
1 answer:
Drupady [299]3 years ago
6 0

Answer:

D: 8

Step-by-step explanation:

7 + (2 + 6) ^2 ÷ 4 ⋅ (1/2)^4

According to PEMDAS

We to parentheses first

7 + (8)^ 2 ÷ 4 ⋅ (1/2)^4

Then we do exponents

7 + 64 ÷ 4 ⋅ (1/16)

The multiply and divide from left to right

7+64 ÷ 4 ⋅ (1/16)

7+16 ⋅ (1/16)

Then add and subtract from left to right

7+1

8

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27p4-27p2-12p+12/p

Step-by-step explanation:

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Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

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Answer:

The answer is A.

Tip:

Press control + U to find the answers in Forms

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