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2 years ago

The figure shown has two parallel lines cut by a transversal:

Mathematics

1 answer:

dangina [55]2 years ago

5 0

Answer: $\angle 2$

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Determine what type of observational study is described. Explain. Vitamin D is important for the metabolism of calcium and expos

ICE Princess25 [194]

Answer:

thats too long to read but good luck!

Step-by-step explanation:

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3 years ago

Please help me out The equation 9(w + 7) = 27 solved in several steps below. For each step, choose the reason that best justifie

Vaselesa [24]

9w+63=27

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3 years ago

What is the difference of 8 times a number r and 12?

lapo4ka [179]

Answer:

The difference of 8 times a number r and 12 is solved for a number r and is given by $r=\frac{3}{2}$

Step-by-step explanation:

Given problem is the difference of 8 times a number r and 12

It can be written as

$(8\times r)-12$

To simplify the above expression :

8r-12

$8r-12=0$

$8r-12+12=0+12$ ( using addition property of equality )

$8r=12$

$8r(\frac{1}{8})=12(\frac{1}{8})$ (using division property of equality )

$r=\frac{3}{2}$

Therefore $r=\frac{3}{2}$

Therefore the difference of 8 times a number r and 12 is solved for a number r and is given by $r=\frac{3}{2}$

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3 years ago

Which table shows a constant rate of change?

Dafna1 [17]

Answer:

B. 2nd Table

Step-by-step explanation:

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3 years ago

Read 2 more answers

3. Let U and V be subspaces of a vector space W. Prove that their intersection UnV is also a subspace of W

kenny6666 [7]

Answer: The proof is done below.

Step-by-step explanation: Given that U and V are subspaces of a vector space W.

We are to prove that the intersection U ∩ V is also a subspace of W.

(a) Since U and V are subspaces of the vector space W, so we must have

0 ∈ U and 0 ∈ V.

Then, 0 ∈ U ∩ V.

That is, zero vector is in the intersection of U and V.

(b) Now, let x, y ∈ U ∩ V.

This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.

Since U and V are subspaces of U and V, so we get

x + y ∈ U and x + y ∈ V.

This implies that x + y ∈ U ∩ V.

(c) Also, for a ∈ R (a real number), we have

ax ∈ U and ax ∈ V (since U and V are subspaces of W).

So, ax ∈ U∩ V.

Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have

x + y and ax ∈ U ∩ V.

Thus, U ∩ V is also a subspace of W.

Hence proved.

7 0

3 years ago