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Nostrana [21]
3 years ago
5

sub to my y o t u b e. my. y o u t u b e. is stiko9 the vidoe tha says oh no primo or small skate clips or slomo kickflip is my

video just click on the video and sub thanks help me hit 20 subs.
Mathematics
2 answers:
julia-pushkina [17]3 years ago
5 0

Answer:

i said im from brainly on slo mo kickflip

Step-by-step explanation:

ArbitrLikvidat [17]3 years ago
4 0

Answer:

yay free points thanks

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Debby needed one-third of a cup of water for 1 flower if she had nine flowers how many cups wold she need
4vir4ik [10]
To solve this problem, you must multiply the 1/3 cup of water needed for one flower, by 9. 
1          9            9
---  *   -----   =   ----  =  3 cups
3           1           3
5 0
3 years ago
Veronica bought an ICEE for $6 at the movies. She also bought some hot dogs for $3 each. Veronica did not want to spend more tha
Scorpion4ik [409]

27 dollars minus 6 for the ice leaves 21 for hot dogs.. hot dogs are 3 each and 21 dollars left it would be 21 divided by 3 which equals 7

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3 years ago
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
A tv that normally cost $800
sukhopar [10]

what is the question it might be an Sony-75"

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3 years ago
PLEASE HELP!!!!!!!!!​
STatiana [176]

Answer:

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