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Nookie1986 [14]

2 years ago

9

According to the fundamental theorem of algebra, how many roots does the polynomial f(x)=x4+3x2+7 have over the complex numbers,

and counting roots with multiplicity greater than one as distinct? (i.e f(x)=x2 has two roots, both are zero).

1 answer:

Finger [1]2 years ago

8 0

ANSWER

4 roots.

EXPLANATION

The given polynomial function is

$f(x) = {x}^{4} + 3 {x}^{2} + 7$

This is a fourth degree polynomial.

According to the Fundamental Theorem of Algebra, an nth degree polynomial has n roots.

This roots include both real and complex roots.

Also repeated roots or roots with multiplicity greater than one are counted as distinct.

Since the given polynomial function is a fourth degree polynomial, the Fundamental Theorem of Algebra, says that this polynomial has 4 roots.

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Answer:

24

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3 years ago

2x+2y=8 in function form step by step please?

nasty-shy [4]

You could get two different functions out of this:

<u>2x + 2y = 8</u>

1). <u> 'y' as a function of 'x'</u>

Subtract 2x from each side: 2y = -2x + 8

Divide each side by 2 : <em> y = -x + 4</em>

2). <u> 'x' as a function of 'y' :</u>

Subtract 2y from each side: 2x = -2y + 8

Divide each side by 2 : <em> x = -y + 4</em>

Those two functions look the same, but that's just because the original
equation given in the problem was so symmetrical. In general, they're
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7 0

2 years ago

Read 2 more answers

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

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2 years ago

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He would need a 93% on his fourth exam.

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2 years ago