What is (6x10^6)(4x10^-1) ?

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

Multiply the 6 and 4 together. 6*4 = 24

The next part is a bit tricky. Add the powers on the base 10. Keep the base 10.

10^(6 - 1) = 10^5

Now put the two parts together. 24 * 10^5

The number in front of the 10 must be between 1 and 10.

So 24 can be written as 24 = 2.4 * 10^1

2.4 * 10^1 * 10^5 = 2.4 * 10^(5 + 1) = 2.4 * 10^6


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which of the following statistic measures the most frequently occuring value in a set of data A.) range B.) mean C.) mode D.) median ​

Answers

Answer:

This is mode defined

Step-by-step explanation:

Enter the range of values for x

Answers

Greetings from Brasil...

See the attached figure. The smaller the θ angle, the smaller the AB side will be. If the angle θ = 90º, then AB = 25. As θ < 90, then AB < 25

5X - 10 < 25

5X < 25 + 10

X < 35/5

X < 7

The AB side can be neither zero nor negative. So

5X - 10 > 0

5X > 10

X > 10/5

X > 2

2 < X < 7

Suppose that the value of a stock varies each day from $12.82 to $33.17 witha uniform distribution.
Find the third quartile; 75% of all days the stock is below what value?

Answers

Answer:

$28.08

Step-by-step explanation:

$33.17 - $12.82 = $20.35

75% of $20.35 = $15.26

$15.26 + $12.82 = $28.08

Can anyone help me plzzzzz

Answers

Step-by-step explanation:

cmon man its a blank screen

Show that W is a subspace of R^3.

Answers

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} $.

Solbe for x : x-1/x+3x+3=0​

Answers

Answer:

x=1/4 or x=-1

Step-by-step explanation:

x−

1

x

+3x+3=0

4x2+3x−1

x

=0

Step 1: Multiply both sides by x.

4x2+3x−1=0

(4x−1)(x+1)=0(Factor left side of equation)

4x−1=0 or x+1=0(Set factors equal to 0)

x=

1

4

or x=−1

Check answers. (Plug them in to make sure they work.)

x=

1

4

(Works in original equation)

x=−1(Works in original equation)

Answer:

x=

1

4

or x=−1