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Olenka [21]
3 years ago
12

Answer answer it it it Answer answer it it it

Mathematics
2 answers:
dexar [7]3 years ago
7 0

Answer: C

Step-by-step explanation:

6+3=9

9+3=2

ryzh [129]3 years ago
7 0

Answer: A

Step-by-step explanation:

6+3=9

9+3=12

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What is the expanded form of the decimal 444.3? A) (4 x 100) + (4 x 10) + (4 x 1) + (3 x 1 10 ) B) (4 x 100) + (4 x 10) + (4 x 1
Tpy6a [65]

444.3 = (4 x 100) + (4 x 10) + (4 x 1) + (3 x 1 /10 )

Answer

A. (4 x 100) + (4 x 10) + (4 x 1) + (3 x 1 /10 )


6 0
2 years ago
Read 2 more answers
Answer for brainliest
Tcecarenko [31]
Answer:
If this is what your looking for, the answer may be:
-4y + 12
Hope this helps!
5 0
2 years ago
Pls help asap, find the height of the cylinder and round to the nearest tenth
Elena L [17]

Answer:

h = 2.4

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h  where r is the radius and h is the height

271.4 = pi ( 6)^2 *h

271.4 = 36* pi* h

Letting 3.14 = pi

271.4 = 113.04 h

Divide each side by 113.04

271.4 /113.04 = 113.04h/113.04

2.424805379 = h

Rounding to the nearest tenth

2.4 = h

OR if we use the pi button

271.4 = 36 * pi *h

Divide each side by 36 pi

271.4/36pi = h

h=2.3997

Rounding to the nearest tenth

h =2.4

3 0
2 years ago
Extended rotation rules
Westkost [7]
Ehjaklopiuftdhsabmzcbs
8 0
3 years ago
Read 2 more answers
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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