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irina [24]
3 years ago
7

Please I need help?!!!

Mathematics
1 answer:
Grace [21]3 years ago
4 0

Answer: Thats easy dude come on you know this

Step-by-step explanation:

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Mayra's monthly salary is $4,540 plus she earns an 8% commission on her sales each month. Mayra's sales for January were $30,000
lesantik [10]
Work out 8% of 30000

10% of 30000 is 3000

1% of 30000 is 300
2% of 30000 is 300x2=600

Therefore 10%-2%= 3000-600=2400

It therefore follows that Mayra earned 4540 + 2400 = 6940 for January

7 0
3 years ago
If n is an integer, which conjecture is not true about 2n– 1? A. 2n– 1 is odd if n is positive. B. 2n– 1 is always even. C. 2n–
VladimirAG [237]
B.



Let's simply look at each conjecture and determine if it's true or false.



A. 2n– 1 is odd if n is positive: Since n is an integer, 2n will always be even. And an even number minus 1 is always odd. Doesn't matter if n is positive or not. So this conjecture is true.



B. 2n– 1 is always even: Once again, 2n will always be even. So 2n-1 will always be odd. This conjecture is false.



C. 2n– 1 is odd if n is even: 2n is always even, so 2n-1 will always be odd, regardless of what n is. So this conjecture is true.



D. 2n– 1 is always odd: 2n will always be even. So 2n-1 will always be odd. Once again, this conjecture is true.



Of the 4 conjectures above, only conjecture B is false. So the answer is B.
6 0
3 years ago
<img src="https://tex.z-dn.net/?f=2a%20%3D%204a%20%2B%2022" id="TexFormula1" title="2a = 4a + 22" alt="2a = 4a + 22" align="absm
Vera_Pavlovna [14]
A is -11
Is that what you're trying to find
7 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
What is the distance between the points H(-5,9) and J(1,6)?
Ahat [919]

Answer:The answer would be b

Step-by-step explanation:

8 0
3 years ago
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