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Ostrovityanka [42]
3 years ago
5

Based on the type of equations in the system, what is the greatest possible number of solutions? StartLayout Enlarged left-brace

1st Row x squared + y squared = 9 2nd row 9 x + 2 y = 16 EndLayout
Mathematics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

2

Step-by-step explanation:

Given the system of equations:

x^2+y^2=9\\9x+2y=16

Comparing x^2+y^2=9 with the general standard equation of a circle (x-h)^2+(y-k)^2=r^2.

The first equation is an <u>equation of a circle centred</u> at (0,0) with a Radius of 3.

The second equation 9x+2y=16 is a <u>straight line equation.</u>

A straight line can only intersect a circle at a maximum of 2 points.

Therefore the greatest possible number of solutions to the equations in the system is 2.

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Answer: A. It would be shifted up.

Step-by-step explanation: the only thing that is changing in the two equations is the last number. The last number is not there in the second equation because the y-intercept is 0. The y-intercept in the first equation is -2 and shifts up 2 to y-intercept if 0. Therefore, your answer would be A. It would shift up.

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What is 7285 divided by 4
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What are the possible rational roots of the polynomial equation?<br><br> 0=2x7+3x5−9x2+6
RoseWind [281]

Answer: \pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0

Where:

- The coefficients are integers.

- a_n is the leading coeffcient (a_n\neq 0)

- a_0 is the constant term a_0\neq 0

Every rational root of the polynomial is in the form:

\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}

For the case of the given polynomial:

2x^7+3x^5-9x^2+6=0

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

 Then, by Rational Roots Test we get the possible rational roots of this polynomial:

\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}

5 0
3 years ago
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