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padilas [110]
3 years ago
7

Round to the merest hundredth389.993

Mathematics
1 answer:
kvasek [131]3 years ago
3 0
What's a hundredth?

it's like this:

0.01

so where is the hundred place in this number?

389.9<em>9</em>3

it's the 9 that i made boldface

so to round to this number we delete everything after it, asking ourselves if the number after it is 4 or less? yes! so we can just delete it without changing anyway


389.99

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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
5 3⁄10 hectometers = meters
Morgarella [4.7K]

Answer:

\large\boxed{5\dfrac{3}{10}\ hectometers=530\ meters}

Step-by-step explanation:

PREFIXES\\\\mega\ (M)=10^6\\\\kilo\ (k)=10^3\\\\hecto\ (h)=10^2\\\\deko\ (da)=10\\\\decy\ (d)=10^{-1}=0.01\\\\centi\ (c)=10^{-2}=0.01\\\\mili\ (m)=10^{-3}=0.001\\\\mikro\ (\mu)=10^{-6}=0.000001

5\dfrac{3}{10}=5.3\\\\1\ hectometers=100\ meters\qquad(1\ hm=100m)\\\\5\dfrac{3}{10}\ hm=5.3\ hm=5.3\cdot100\ m=530\ m

8 0
3 years ago
Harry bought 25 pens for 1.00. Find the unit rate
lara31 [8.8K]

Answer:

4 cents

Step-by-step explanation:

1.00 divied by 25 =0.04 aka 4 cents

6 0
3 years ago
Read 2 more answers
What is the length of AB?
kramer

Answer:

We are given to find the length of side AB. We know that in a triangle, if two angles have equal measures, then the sides opposite to them are equal in length. Thus, the length of side AB is 6 units.

Hope this helps!!

3 0
3 years ago
Tony’s fish weighs five pounds more than three times the weight of Mary’s fish. Let t represent the weight of Tony’s fish, and l
Temka [501]

t = 5 + 3m is the required expression that represents weight of Tony fish

<h3><u>Solution:</u></h3>

Let "t" represent the weight of Tony’s fish, and let "m" represent the weight of Mary’s fish

To find: expression that represents the weight of Tony's fish

According to given information,

Tony’s fish weighs five pounds more than three times the weight of Mary’s fish

Here the word "times" represents multiplication and "more than" represents addition

Weight of Tony fish = 5 + three times the weight of Mary’s fish

Weight of Tony fish = 5 + 3(m)

t = 5 + 3m

Thus the required expression is found out

6 0
4 years ago
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