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AVprozaik [17]
3 years ago
5

What is 8x + 16 + xy + 2y as factor each expression completely.

Mathematics
1 answer:
liraira [26]3 years ago
8 0

Answer:

( x + 2 ) ( 8 + y )

Step-by-step explanation:

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irina [24]

Answer: At x = 2 the graph crosses the x-axis

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8.7, 8.9, 9.0 is the answer
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3 years ago
Select the graph that would represent the best presentation of the solution set for Ixl < 5.
Lelu [443]

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5.D1.3 select from among a variety of graphs, including stacked-bar graphs, the type of graph best suited to represent various sets of data; display

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7 0
2 years ago
If f(2)=9 and f(-1)=14, write a linear function that fits this scenario. Be sure to use proper notation. HELP!!
olya-2409 [2.1K]

Answer:

The required linear function is: f(x) = $ \frac{-5}{3}x + \frac{37}{3} $

Step-by-step explanation:

We are given that f(x) is a linear function and it takes the value 9 when x = 2 and 14 when x = -1.

Now the general form of any linear function is: f(x) = ax + b.

Substituting these values in the general form we get:

f(2) = 9  = 2a + b

f(-1) = 14 = -a + b

Solving these two equations we get:

b = 37/3

Substituting this in the second equation to find 'a'.

a = -5/3

Therefore, the function f(x) = $ \frac{-5}{3} $x + $ \frac{37}{3} $.

4 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
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