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3 years ago

25 points+brainliest!

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

Answer:

an= 10n+90

a10= 190

Serggg [28]3 years ago

5 0

Answer:

$\displaystyle a_n=100+10(n-1)$

$\displaystyle a_{10}=190$

Step-by-step explanation:

We have the arithmetic sequence:

100, 110, 120, 130...

And we want to determine the equation for the nth term of the sequence.

The standard form for the nth term of an arithmetic sequence is given by the formula:

$\displaystyle a_n=a+d(n-1)$

Where $a_n$ is the nth term, $a$ is the first term, and $d$ is our common difference.

From the sequence, we can see that the first term $a$ is 100.

Also, each term is +10 the previous term. Hence, our common difference $\displaystyle d$ is +10.

Therefore, our equation is:

$\displaystyle a_n=100+10(n-1)$

To find the 10th term, we can substitute 10 for n and evaluate. Hence:

$\displaystyle a_{10}=100+10(10-1)$

Evaluate:

$\displaystyle a_{10}=100+10(9)=100+90=190$

Therefore, the 10th term is 190.

You might be interested in

5(x-10)=30-15x Ox=1 Ox=4 Ox=5 Ox=8

makvit [3.9K]

Answer:

x = 4

Step-by-step explanation:

5x-50=30-15x

5x +15x =30+50

20x= 80

Divide both sides of the equation by 20

x = 4

I hope it helps

7 0

3 years ago

An n × n matrix B has characteristic polynomial p(λ) = −λ(λ − 3) 3 (λ − 2) 2 (λ + 1). Which of the following statements is false

asambeis [7]

Answer:

Only d) is false.

Step-by-step explanation:

Let $p=p(\lambda)=\lambda(\lambda-3)^3 (\lambda-2)^2 (\lambda+1)$ be the characteristic polynomial of B.

a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)

b) Remember that $p(\lambda)=\det(B-\lambda I)$ . 0 is a root of p, so we have that $p(0)=\det(B-0 I)=\det B=0$ .

c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.

d) det(B)=0 by part c) so B is not invertible.

e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.

8 0

3 years ago

Solve and get brainliest!! please be sure of your answer!!!!

laila [671]

Answer:

hope this helps youuuuu