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jarptica [38.1K]
3 years ago
7

1,000 kilograms equals 1 _______?

Mathematics
1 answer:
krok68 [10]3 years ago
5 0
1000 kg
1000 kilo = 1 Mega = 1 M

so
1000 kg = 1 Mg
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Estimate 1299+13,522 +6424 by first rounding each number to the nearest thousand.
ohaa [14]

Answer:

1299+13522+6424=21245

round off to the nearest thousand

2<u>1</u>,245

the nearest thousand is 1,000

the number beside it is 200

according to the rule to round off it has to be above 5 it is not

so drop off the 2 and add zero to the end

now your answer is 21,000

Step-by-step explanation:

hope this helps good luck!

7 0
2 years ago
Read 2 more answers
What is the slope of the line represented by the equation y + 3 = -4(x-5)?
garik1379 [7]

Answer:

-4

Step-by-step explanation:

if you graph it, we can find out two points are 0,17 and 1,13

we can find the slope from rise/run

6 0
2 years ago
Easy answer!!!! Will give brainliest
pochemuha

Answer:

A and B

Step-by-step explanation:

6 0
3 years ago
Given the equation square root of quantity 3x plus 6 end quantity equals 3, solve for x and identify if it is an extraneous solu
MissTica

Step-by-step explanation:

3x+6=3

-6 -6

3x =-3

÷3 ÷3

x= -1

4 0
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
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