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notsponge [240]
2 years ago
14

What are the solutions to the following system of equations? x + y = 3 y = x2 − 9 (3, 0) and (1, 2) (−3, 0) and (1, 2) (3, 0) an

d (−4, 7) (−3, 0) and (−4, 7)
Mathematics
1 answer:
OLga [1]2 years ago
3 0

The solutions for the given system of equations are:

(3, 0), (-4, 7).

<h3>How to solve the given system of equations:</h3>

Here we have the system:

x + y = 3

y = x² - 9

To solve this, we can replace the second equation into the first one, so we get:

x + (x² - 9) = 3

Now we can solve this quadratic equation for x, we need to solve:

x² + x - 12 = 0.

The solutions are given by Bhaskara's formula:

x = \frac{-1 \pm \sqrt{1^2 - 4*1*(-12)} }{2*1} \\\\x = \frac{-1 \pm 7 }{2}

Then the two solutions are:

x = (-1 + 7)/2 = 3

x = (-1 - 7)/2 = -4

To get the y-values correspondent, we can evaluate the linear equation in these two values:

y = 3 - x.

For x = 3:

y = 3 - 3 = 0

For x = -4:

y = 3 + 4 = 7

Then the two solutions are: (3, 0), (-4, 7).

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

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I need shown work with numbers or an explanation if need.
sveticcg [70]

Let us denote the semi arcs as congruent angles. This means that angles FEJ and EFJ are congruent (That is, they have the same measure). Since angles FEJ and EFJ have the same measure, this implies that sides EJ and FJ are equal. Since angles EJK and FJH are supplementary angles to angle EJF, this implies that EJK and FJH have the same measure.

Using the Angle Side Angle (SAS) criteria, we determine that triangles EKJ and triangle FJH are congruent. This implies that sides EK and FH are equal and that angles EKJ and FHJ are congruent. Note that angle EKJ is the same as EKF and that FHJ is the same as FHE.

Once again, since angles EKF and FHJ are congruent, and angle EKD is supplementary to the angle EKJ when angle FHG is supplementary to angle FHJ, then we have that angles EKD and angle FHG are congruent.

Using again the SAS criteria, we determine that triangles EKD and FHG are congruent.

From this reasoning, we have proved the following facts:

Triangle DEK is congruent to triangl GFH

Angle EKF is congruent to angle FHE

Segment EK is the same as segment FH

4 0
1 year ago
Write an equation of an ellipse in standard form with the center at the origin and a height of 12 units and width of 19 units.
gladu [14]

Answer:

The equation of the ellipse in standard form is 4x²/361 + y²/36 = 1

Step-by-step explanation:

* Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with center (0 , 0 )  

  and major axis parallel to the x-axis is x²/a² + y²/b² = 1  

# a > b  

- The length of the major axis is 2a  

- The coordinates of the vertices are ( ± a , 0 )  

- The length of the minor axis is 2b  

- The coordinates of the co-vertices are ( 0 , ± b )  

- The coordinates of the foci are ( ± c , 0 ) , where c ² = a ² − b²  

* Lets solve the problem

∵ The center of the ellipse is (0 ,0)

∵ Its width is 19 units

∴ The length of the major axis is = 19

∴ 2a = 19 ⇒ divide both sides by 2

∴ a = 19/2 ⇒ ∴ a² = 361/4

∵ Its height is 12 units

∴ The length of the minor axis is = 12

∴ 2b = 12 ⇒ divide both sides by 2

∴ b = 12/2 = 6 ⇒ ∴ b² = 36

- Lets write the equation in standard form

∵ The equation is x²/a² + y²/b² = 1

∴ x²/(361/4) + y²/36 = 1 ⇒ simplify it

∴ 4x²/361 + y²/36 = 1

* The equation of the ellipse in standard form is 4x²/361 + y²/36 = 1

7 0
3 years ago
Help Please!!! 20 points!
Mazyrski [523]
-6/16.................brainilyst???? plz シ
3 0
3 years ago
Plssss help meeee!!!
pogonyaev
The answer is c b/c you multiply
3 0
3 years ago
Read 2 more answers
Does 23^-1 (mod 1000) exist? If yes solve it.
sweet [91]

Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).

Let <em>x</em> be the inverse. Then <em>x</em> is such that

23<em>x</em> ≡ 1 (mod 1000)

Use the Euclidean algorithm to solve for <em>x</em> :

1000 = 43×23 + 11

23 = 2×11 + 1

→   1 ≡ 23 - 2×11   (mod 1000)

→   1 ≡ 23 - 2×(1000 - 43×23)   (mod 1000)

→   1 ≡ 23 - 2×1000 + 86×23   (mod 1000)

→   1 ≡ 87×23 - 2×1000 ≡ 87×23   (mod 1000)

→   23⁻¹ ≡ 87   (mod 1000)

3 0
3 years ago
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