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SashulF [63]
3 years ago
7

What is the answer to this problem to check my work

Mathematics
1 answer:
Valentin [98]3 years ago
6 0
The equation describes a circle of radius 2 centered at (x, y) = (7, -8). In standard form, the equation would be
.. (x -7)^2 +(y +8)^2 = 4

A graph of it can be seen here. https://www.desmos.com/calculator/14zlfrtoa1
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Josh walks 18 dogs each week today he is walking 1/3 of the dogs how many dogs is he walking tiday?
stepan [7]
He is walking 6 dogs.
18÷3=6
7 0
3 years ago
Read 2 more answers
What is the ratio of 16 cm and 448m
damaskus [11]
There are 100 cm in a meter so 16 cm is equal to 1600/48
4 0
3 years ago
Read 2 more answers
It costs $12 to get into the San Diego County Fair and $1.50 per ride.
yan [13]

Answer:

8

Step-by-step explanation:

(24-12)/1.50=8

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
Please help me ASAP!!!! :) I’d appreciate it
Amiraneli [1.4K]

Answer:

5×(-4) = -20

(-1) × 4 = -4

Step-by-step explanation:

Here it' sayin when a positive number ( it may be odd or even ) gets multiplied with an even number ( it may be positive or negative ) , the product will be positive number.

5×(-4) = -20 .Here 5 is a positive number & 4 is a negative even number but the product -20 is a negative number. So it contradicts the statement.

(-1) × 4 = -4 . Here -1 is a negative number & 4 is a positive even number but the product -4 is a negative number. So it contradicts the statement.

5 0
3 years ago
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