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QveST [7]
3 years ago
9

Write the expression as a product of polynomials:

Mathematics
1 answer:
Sonja [21]3 years ago
3 0
(b+c)(x+3)
I hope this is true
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What is the possible rational roots for this polynomial 6x³+5x²+10x+3=0
8090 [49]

Answer:

See below.

Step-by-step explanation:

Using the Rational Roots Theorem:

Factors of 3:   1, 3  ( = p).

Factors of 6:  1,2,3,6 ( = q).

Possible real  roots are a 1 or 3 from  p / q = +/- 1/1, +/- 3/1 , +/-1/2,  +/- 1/3 , +/- 1/6, +/- 3/2.

3 0
3 years ago
Round 2.71828 to the hearest thousandth.
Colt1911 [192]

Answer:

2.718

Step-by-step explanation:

The thousandths place is 3 to the right of the decimal (the 8).

Since the next number is 2, the number stays the same and does not round up.

Hope this helps :)

4 0
3 years ago
Which pair shows equivalent expressions?
Yuliya22 [10]
C is the answer to this question.
4 0
2 years ago
Read 2 more answers
1 point) Are the functions f,g, and h given below linearly independent? f(x)=e3x+cos(5x), g(x)=e3x−cos(5x), h(x)=cos(5x). If the
Fynjy0 [20]

Answer:

Functions are linearly dependent (are not linearly independent.)

Step-by-step explanation:

Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the <em>only solution</em> to the equation

\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I

is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.

Now, let f(x), g(x) and h(x) be the functions:

f(x)=e^{3x}+\cos(5x),

g(x)=e^{3x}-\cos(5x),

h(x)=\cos(5x).

Then, letting α = 1, β= -1 and ω = -2, we see that:

\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.

Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.

8 0
3 years ago
Z +9&gt; 22<br> Simplify your answer as much as possible.
Aleonysh [2.5K]

Answer:

z > 13

Step-by-step explanation:

Subtract 9 from 22.

z > 13

6 0
3 years ago
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