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3 years ago

What is 1.53 to the nearest tenth

Mathematics

2 answers:

Marat540 [252]3 years ago

7 0

1.53 rounded to the nearest tenth is 1.5

Natali [406]3 years ago

5 0

Answer:

1.50

Step-by-step explanation:

The tenths place is the number to the exact right of the decimal, in this case it would be the 5. To round, you need to look to the number to the right of it, in this case it would be the 3.

There's this rule, there's also a rhyme, but I can't remember it atm, anyways it's 5 or higher you round up, 4 or lower you round down. So because it's a 3 you round down, which bascially means it stays the same.

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Why is the fraction 6/10 not written 0.06?

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3 years ago

Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

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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.

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