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SpyIntel [72]
3 years ago
10

What are the coordinates of vertex ?

Mathematics
1 answer:
Step2247 [10]3 years ago
8 0

Answer:

(0, -8)

Step-by-step explanation:

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PLZ HELP ME
True [87]

Answer:

∠ 1 = 20°, ∠ 2 = 160°

Step-by-step explanation:

∠ 1 and ∠ 2 are adjacent angles and are supplementary, that is sum to 180°

x - 12 + 5x = 180 , that is

6x - 12 = 180 ( add 12 to both sides )

6x = 192 ( divide both sides by 6 )

x = 32

Thus

∠ 1 = x - 12 = 32 - 12 = 20°

∠ 2 = 5x = 5 × 32 = 160°

5 0
3 years ago
Chrissy had 4 gallons of gas in her tank when she arrived at the gas station. She pumped gas into her car at a rate of <img src=
vichka [17]
It would be 15 I think : )
8 0
3 years ago
How many factors does 18 have
Zepler [3.9K]
Factors of 18: 1, 2, 3, 6, 9, 18

So, it has 6 factors.
8 0
3 years ago
Show that W is a subspace of R^3.
musickatia [10]

Answer:

Check the two conditions of Subspace.

Step-by-step explanation:

If W is a Subspace of a vector space, V then it should satisft the following conditions.

1) The zero element should be in W.

Zero element can be different for different vector spaces. For examples, zero vector in $ \math{R^2} $ is (0, 0) whereas, zero element in $ \math{R^3} $ is (0, 0 ,0).

2) For any two vectors, $ w_1 $ and $ w_2 $ in W, $ w_1 + w_2 $ should also be in W.

That is, it should be closed under addition.

3) For any vector $ w_1 $ in W and for any scalar, $ k $ in V, $ kw_1 $ should be in W.

That is it should be closed in scalar multiplication.

The conditions are mathematically represented as follows:

1) 0$ \in $ W.

2) If $ w_1 \in W; w_2 \in W $ then $ w_1 + w_2 \in W $.

3) $ \forall k \in V, and \hspace{2mm} \forall w_1 \in W \implies kw_1 \in W

Here V = $ \math{R^3} $ and W = Set of all (x, y, z) such that $ x - 2y + 5z = 0 $

We check for the conditions one by one.

1) The zero vector belongs to the subspace, W. Because (0, 0, 0) satisfies the given equation.

i.e., 0 - 2(0) + 5(0) = 0

2) Let us assume $ w_1 = (x_1, y_1, z_1) $ and $ w_2 = (x_2, y_2, z_2) $ are in W.

That means: $ x_1 - 2y_1 + 5z_1 = 0 $ and

$ x_2 - 2y_2 + 5z_2 = 0 $

We should check if the vectors are closed under addition.

Adding the two vectors we get:

$ w_1 + w_2 = x_1 + x_2 - 2(y_1 + y_2) + 5(z_1 + z_2) $

$ = x_1 + x_2 - 2y_1 - 2y_2 + 5z_1 + 5z_2 $

Rearranging these terms we get:

$ x_1 - 2y_1 + 5z_1 + x_2 - 2y_2 + 5z_2 $

So, the equation becomes, 0 + 0 = 0

So, it s closed under addition.

3) Let k be any scalar in V. And $ w_1 = (x, y, z) \in W $

This means $ x - 2y + 5z = 0 $

$ kw_1 = kx - 2ky + 5kz $

Taking k common outside, we get:

$ kw_1 = k(x - 2y + 5z) = 0 $

The equation becomes k(0) = 0.

So, it is closed under scalar multiplication.

Hence, W is a subspace of $ \math{R^3} $.

7 0
3 years ago
What is 15 1/8 minus 1 3/5?
d1i1m1o1n [39]
I hope this helps you

8 0
4 years ago
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