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Marianna [84]
3 years ago
10

5(x+1)=4x+21 solve please

Mathematics
1 answer:
REY [17]3 years ago
7 0
The answer: 16

You have to move the terms then solve the problem
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If f(3) = 9, what is f(1)<br>​
Vaselesa [24]

Answer:

3

Step-by-step explanation:

3*3=9

1*3=3

8 0
3 years ago
Read 2 more answers
Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se
defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

p_{1}(t) = 1, p_{2}(t)= t^{2} and p_{3}(t) = 3 + 3\cdot t:

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0

(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0

The following system of linear equations is obtained:

\alpha_{1} + 3\cdot \alpha_{3} = 0

\alpha_{2} = 0

\alpha_{3} = 0

Whose solution is \alpha_{1} = \alpha_{2} = \alpha_{3} = 0, which means that the set of vectors is linearly independent.

p_{1}(t) = t, p_{2}(t) = t^{2} and p_{3}(t) = 2\cdot t + 3\cdot t^{2}

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0

(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0

The following system of linear equations is obtained:

\alpha_{1}+2\cdot \alpha_{3} = 0

\alpha_{2}+3\cdot \alpha_{3} = 0

Since the number of variables is greater than the number of equations, let suppose that \alpha_{3} = k, where k\in\mathbb{R}. Then, the following relationships are consequently found:

\alpha_{1} = -2\cdot \alpha_{3}

\alpha_{1} = -2\cdot k

\alpha_{2}= -2\cdot \alpha_{3}

\alpha_{2} = -3\cdot k

It is evident that \alpha_{1} and \alpha_{2} are multiples of \alpha_{3}, which means that the set of vector are linearly dependent.

p_{1}(t) = 1, p_{2}(t)=t^{2} and p_{3}(t) = 3+3\cdot t +t^{2}

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0

(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0

The following system of linear equations is obtained:

\alpha_{1}+3\cdot \alpha_{3} = 0

\alpha_{2} + \alpha_{3} = 0

3\cdot \alpha_{3} = 0

Whose solution is \alpha_{1} = \alpha_{2} = \alpha_{3} = 0, which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

4 0
3 years ago
Will give crown help
jasenka [17]

Answer:

D is a right angle

B is an accute angle

A and C are obtuse

4 0
3 years ago
An employee at a store makes $6 an hour. If she has made $237.38 in a week, how
nalin [4]

Answer:40 hours

Step-by-step explanation:divide 237.36 by 6 and you will get 39.5633

so basically what you're going to do is round it up and your answer will be 40

8 0
3 years ago
Evalute b (caret) 2-5c b=15 and c =9
Advocard [28]

Answer: look at the 4 picture for the answer’s

Question: 2-5c b=15

Question: 2-5c c=9

Step-by-step explanation: hope this help:)

5 0
3 years ago
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