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zhuklara [117]
3 years ago
11

Need help with this problem h+-3=4 please

Mathematics
2 answers:
Crazy boy [7]3 years ago
8 0

Answer:

h=7

Step-by-step explanation:

1) Add three to both sides

2) You should get h=7

Anna71 [15]3 years ago
7 0

Answer:

h=7

Step-by-step explanation:

h+(-3)=4

may be rewritten as

h-3=4

as adding a negative is the same as subtracting a positive.

To solve, add 3 to both sides.

h-3=4\\h=7

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Solve the inequalities; <br><br><br>1. |4x + 5| + 2 ≥ 10<br><br><br>2. |2x + 3| + 6 &gt; 0
slamgirl [31]

Answer:

Number two: x > -9/2, x < 3/2

Number one: x is less than or equal to 3/4, x is greater than or equal to - 13/4

Step-by-step explanation:

Number one:

|4x+5| > 10-2

|4x+5| > 8

4x+5 > 8

-(4x+5) > 8

x > 3/4

x < -13/4

x > 3/4, x < -13/4

(all of these signs are greater than or equal to and less than or equal to)

Number two:

|2x+3| > -6

2x+3 > -6

-(2x+3) > -6

x > -9/2

x < 3/2

x> -9/2, x < 3/2

(all of these signs are correct and not greater than or equal to or less than or equal to)

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3 years ago
A radio cost $45.00 and is on sale for 30% off. What's the final price of the radio after the discount
stira [4]

I think it might be $31.50! Hope it helps1



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Nathan's calculator displays the following: 5.987e-5
vampirchik [111]

Answer:

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What is the value of x?
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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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