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alina1380 [7]
3 years ago
8

Make a list of the possible outcomes from the table.

Mathematics
1 answer:
DochEvi [55]3 years ago
4 0

Answer:

9 unless the regular shirt and trousers dont count, in which case it's 4

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What is the value of x
Anna11 [10]

Answer:

28

Step-by-step explanation:

14x2=28

Therefore x=28

8 0
3 years ago
Factor the quadratic equation below to reveal the solutions. X^2+4x-21=-9
a_sh-v [17]

Answer:

x = 3 and x = -7

Step-by-step explanation:

The given quadratic equation is x^2+4x-21=0. We need to find the solution of this equation.

If the equation is in the form of ax^2+bx+c=0, then its solutions are given by :

x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = 1, b = 4 and c = -21

Plugging all the values in the value of x, such that :

x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a},\dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-4+ \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)},\dfrac{-4- \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)}\\\\x=3, -7

So, the solutions of the quadratic equation are 3 and -7.      

6 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
Help please! I asked this question three times and my session is about to start I really need to submit it.
hodyreva [135]

Answer: 13) 6

I dont know 16 or 17

19) 33

Step-by-step explanation:

5 0
2 years ago
You are starting a savings account for college. You put $1,000 in as your starting balance. You earn simple interest at 10% ever
Brums [2.3K]

Answer:

— Use NerdWallet to find a better bank account. ... Compound interest is simple: It's the interest you earn on both your original deposit and on the ...

Step-by-step explanation:

7 0
2 years ago
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