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3 years ago

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?

Mathematics

2 answers:

RideAnS [48]3 years ago

4 0

Three! -7/9, -1/4, and -4/3

garri49 [273]3 years ago

3 0

Answer with Step-by-step explanation:

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number can be considered a complex number with its imaginary part equal to zero.

Hence, the given polynomial equation has atleast 1 real root and atmost 3 real roots

(9x + 7)(4x + 1)(3x + 4) = 0

On solving this, we get

9x+7=0 or 4x+1=0 or 3x+4=0

i.e. x= -7/9 or x= -1/4 or x= -4/3

As we can see all the roots are real

Hence, number of real roots are:

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Let X be from a geometric distribution with probability of success p. Given that P(X > y) = (1 ???? p)y for any positive inte

nevsk [136]

Full Question

Let X be from a geometric distribution with probability of success p.

Given that P ( X > a + b | X > a ) = q ^ { b } = P ( X > b ) for any positive integer x.

Show that for positive integers a and b, P(X > a + b | X > a) = P(X > b).

Answer:

P(X > a + b | X > a) = P(X > b)

Proved --- See Explanation

Step-by-step explanation:

Given

P ( X > a + b | X > a ) = q ^ { b } = P ( X > b )

From the above.

We can derive the following

P(X > a), P(X > b) and P(X > a + b)

P(X > a) = q^a

P(X > b) = q^b

P(X > a + b) = q^(a + b)

Using the definition of conditional probability

P(X > a + b | X > a) can be represented by P(X > a + b and X > a)/ P(X>a)

X>a+b and X>a is equivalent to X>a+b since a+b is larger than a

So,

P(X > a + b and X > a)/ P(X>a) can be rewritten as

P(X>a + b)/P(X > a)

Bringing both sides together, we're left with

P(X > a + b | X > a) = P(X>a + b)/P(X > a)

By substituton

P(X > a + b | X > a) = q^(a+b)/q^a

P(X > a + b | X > a) = q^(a + b - a)

P(X > a + b | X > a) = q^(a - a + b)

P(X > a + b | X > a) = q^b

Since P(X > b) = q^b

So,

P(X > a + b | X > a) = P(X > b)

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3 years ago

Evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−4yj+3zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2

Katena32 [7]

WIth the parameterization

$\mathbf r(t)=\langle\sin t,\cos t,t\rangle$

we end up with

$\mathbf f(\mathbf r(t))=\mathbf f(\sin t,\cos t,t)=5\sin t\,\mathbf i-4\cos t\,\mathbf j+3t\,\mathbf k$

We also have

$\mathrm d\mathbf r=\mathbf r'(t)\,\mathrm dt=\langle\cos t,-\sin t,1\rangle\,\mathrm dt$

So in the line integral, we get

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=3\pi/2}\mathbf f(\mathbf r(t))\cdot\mathbf r'(t)\,\mathrm dt$

$=\displaystyle\int_{t=0}^{t=3\pi/2}(9\sin t\cos t+3t)\,\mathrm dt=\frac98(3\pi^2+4)$

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3 years ago

A tuxedo rental shop rents tuxedos with sleeve lengths from 20 inches to 40 inches. The shop says the length of the sleeves shou

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Let the person's arm length be = x

Condition is : sleeve lengths available from 20 inches to 40 inches and the length of the sleeves should be about 1.2 times a person's arm length.

So, the inequality equation becomes:

$20\leq1.2x\leq40$

This gives x= $\frac{20}{1.2}=16.66$ or $16\frac{2}{3}$

or x= $\frac{40}{1.2}$ =33.33 or $33\frac{1}{3}$

Hence, the shop does not provide tuxedos to people with arm length less than 16.66 inches and greater than 33.33 inches.

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nevsk [136]

the answer is 87.75

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