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JulsSmile [24]
1 year ago
8

Find a polynomial of the specified degree that satisfies the given conditions. Degree 4; zeros −1, 1, 7 ; integer coefficients a

nd constant term 14 f(x) =
Mathematics
1 answer:
kirill115 [55]1 year ago
6 0

The polynomial equation is f(x) = (x + 1)(x - 1)(x - 7)(x + 2)

<h3>How to determine the polynomial equation?</h3>

The given parameters are

  • Degree of polynomial = 4
  • Zeros of the polynomial = -1, 1 and 7
  • Constant term = 14

The sum of multiplicities of the polynomial equation must be equal to the degree.

This means that the multiplicity of each zero is 1

The equation of the polynomial is then calculated as

P(x) = (x - zero)^ multiplicity

So, we have

P(x) = (x + 1)(x - 1)(x - 7)(x - a)

Recall that

Constant term = 14

This means that

P(0) = 14

So, we have

(0 + 1)(0 - 1)(0 - 7)(0 - a) = 14

Evaluate

7(0 - a) = 14

Divide by 7

0 - a = 2

So, we have

a = -2

Substitute a = -2 in P(x) = (x + 1)(x - 1)(x - 7)(x - a)

P(x) = (x + 1)(x - 1)(x - 7)(x + 2)

Rewrite as

f(x) = (x + 1)(x - 1)(x - 7)(x + 2)

Hence, the equation of the polynomial equation is f(x) = (x + 1)(x - 1)(x - 7)(x + 2)

Read more about polynomial at

brainly.com/question/17517586

#SPJ1

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Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)
zubka84 [21]
\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } &#10;

This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

\frac{ \sqrt{-49}}{7-2i-4-9i}

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result: 

\frac{ \sqrt{-49} }{3-11i}

Now, the \sqrt{-49} must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to \sqrt{49}  \sqrt{-1}
This means that we get the result 7i for the numerator.

\frac{7i}{3-11i}

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: x^2 - y^2 =(x+y)(x-y)
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

\frac{7i (3+11i)}{(3-11i)(3+11i)}

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how (3-11i)(3+11i)= 3^2 -(11i)^2 therefore, we can finally write:

\frac{7i(3+11i)}{3^2 - (11i)^2 }

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product 7i(3+11i) = 21i+77i^2.
You should know from your classes that i^2 = -1, thefore the numerator simplifies to -77+21i
You can do it as a curious thing, but simplifying yields the result:
\frac{-77+21i}{130}
7 0
3 years ago
4 + {−5 + [−3 + 4 + 2(−7 + 4) + 4] + 2}
scoray [572]

Answer

it is 5 : )

Step-by-step explanation:

8 0
2 years ago
Laura has 30 cupcakes she wants to give each friend 1/6 of her cupcakes how many cupcakes will Each friend get
sdas [7]

Answer:

5 cupcakes

Step-by-step explanation:

=1x 30/ 6

=5

I hope it helps

Thank U

7 0
3 years ago
Read 2 more answers
Is negative square root of 35 rational or irrational?
Vika [28.1K]
Irrational would be the correct answer
8 0
3 years ago
If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Sedbober [7]

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

  • FX(x)=P(X≤x)
  • FY(y)=P(Y≤y).

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

  • t(Y)≤v)
  • Fs(X)(u)=P(s(X)≤u)
  • Ft(Y)(v)=P(t(Y)≤v).

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

7 0
3 years ago
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