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stiv31 [10]
3 years ago
7

{3x+2y=-5 {4x-3y= -18

Mathematics
1 answer:
NemiM [27]3 years ago
3 0

\left\{\begin{array}{ccc}3x+2y=-5&|\cdot3\\4x-3y=-18&|\cdot2\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}9x+6y=-15\\8x-6y=-36\end{array}\right}\qquad\text{add both sides of the equations}\\\\.\qquad17x=-51\qquad|:17\\.\qquad x=-3\\\\\text{put the value of x to the first equation}\\\\3(-3)+2y=-5\\-9+2y=-5\qquad\text{add 9 to both sides}\\2y=4\qquad\text{divide both sides by 2}\\y=2\\\\Answer:\ \boxed{x=-3\ and\ y=2}

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A number y is 5 what times the value of a number X. A line graph the in the coordinate plane represents the relationship between
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Step-by-step explanation:

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Suppose payments were made at the end of each month into an ordinary annuity earning interest at the rate of 2.5%/year compounde
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Answer:

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Step-by-step explanation:

Giving the following information:

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6 0
3 years ago
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professor190 [17]

Answer:

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Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

7 0
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Answer:

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Step-by-step explanation:

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