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photoshop1234 [79]
3 years ago
13

Verify that the vector X is a solution of the given homogeneous system. X' = −1 1 25 1 −1 X; X = −1 5 e−6t/5 For X = −1 5 e−6t/5

, one has

Mathematics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer: Find answer in the attached file

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Factorize (2u+3u)(u+v)-2u+3v​
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Answer:

Step-by-step explanation:

(2u+3u)(u+v)-2u+3v

=2u(u+v)+3u(u+v)-2u+3v

=2u^2+2uv+3u^2+3uv-2u+3v

=5u^2+5uv-2u+3v

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Write the following fractions greater than 1as a sum of two products
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What is one half added to 5 eighths?
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The distance that a object covered in time was measured and recorded in the table below. What equation describes the motion
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Answer:

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Step-by-step explanation:

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Yuri thinks that 3/4 is a root of the following function.
sineoko [7]

Given:

The polynomial function is

q(x)=6x^3+19x^2-15x-28

Yuri thinks that \dfrac{3}{4} is a root of the given function.

To find:

Why \dfrac{3}{4} cannot be a root?

Solution:

We have,

q(x)=6x^3+19x^2-15x-28

If \dfrac{3}{4} is a root, then the value of the function at \dfrac{3}{4} is 0.

Putting x=\dfrac{3}{4} in the given function, we get

q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28

q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28

q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28

q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28

Taking LCM, we get

q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}

q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0

Since the value of the function at \dfrac{3}{4} is not equal to 0, therefore, \dfrac{3}{4} is not a root of the given function.

4 0
3 years ago
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