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notsponge [240]
1 year 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]1 year 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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A newspaper subscription costs $60. Mary Ellen can get a discount of 25% off through her work. How much will Mary Ellen pay for
AysviL [449]

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$45

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The prices of commodities X,Y,Z are respectively x, y, z, rupees per unit. Mr. A purchases 4 units of Z and sells 3 units of X a
liubo4ka [24]

Answer:

(x,y,z)=(1477, 1464, 1437)

Step-by-step explanation:

Consider the selling of the units positive earning and the purchasing of the units negative earning.

<h3>Case-1:</h3>
  • Mr. A purchases 4 units of Z and sells 3 units of X and 5 units of Y
  • Mr.A earns Rs6000

So, the equation would be

3x  +  5y - 4z = 6000

<h3>Case-2:</h3>
  • Mr. B purchases 3 units of Y and sells 2 units of X and 1 units of Z
  • Mr B neither lose nor gain meaning he has made 0₹

hence,

2x   - 3y  +  z = 0

<h3>Case-3:</h3>
  • Mr. C purchases 1 units of X and sells 4 units of Y and 6 units of Z
  • Mr.C earns 13000₹

therefore,

- x    + 4y  +  6z = 13000

Thus our system of equations is

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

<u>Solving </u><u>the </u><u>system </u><u>of </u><u>equations</u><u>:</u>

we will consider elimination method to solve the system of equations. To do so ,separate the equation in two parts which yields:

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\end{cases}\\\begin{cases}2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

Now solve the equation accordingly:

\implies\begin{cases}11x-7y=6000\\-13x+22y=13000\end{cases}

Solving the equation for x and y yields:

\implies\begin{cases}x= \dfrac{223000}{151}\\\\y= \dfrac{221000}{151}\end{cases}

plug in the value of x and y into 2x - 3y + z = 0 and simplify to get z. hence,

\implies z= \dfrac{217000}{151}

Therefore,the prices of commodities X,Y,Z are respectively approximately 1477, 1464, 1437

6 0
2 years ago
(1.8x10^5)divided by(3x10^6) in standard form?
Softa [21]

Answer:

0.06

Step-by-step explanation:

Given data:

1.8(10^{5}) divided by 3×10^{6}

Now,

\frac{(1.8)10^{5} }{3(10^{6}) } \\=0.6(10^{5-6} )\\=0.6(10^{-1} )\\=0.06

Answer will be 0.06

5 0
2 years ago
Pleasee helppp insert two arithnetic means between -5 and 1
Salsk061 [2.6K]

Answer:

a₂ = -3

a₃  = -1

Step-by-step explanation:

We have to  insert two arithmetic means between -5 and 1

Let a₁ and b₂ be the two arithmetic means between -5 and 1

-5, a₂, a₃, 1

Here,

  • a₄ = 1
  • a₁ = -5

We know that the nth term of an Arithmetic sequence

aₙ = a₁ + (n-1)d

a₄ = -5 + (4-1)d

1 = -5 + 3d    

1  + 5 = 3d

3d = 6

Dividing both sides by 2

d = 2

Also

a₂ = a₁ + (2-1)d

a₂ = -5 + d

a₂ = -5 + 2          ∵ d = 2

a₂  = -3

a₃ = a₁ + (3-1)d

a₃  = -5 + 2d

a₃  = -5 + 2(2)   ∵ d = 2

a₃  = -5 + 4

a₃  = -1

Thus,

a₂ = -3

a₃  = -1

Thus, the sequence becomes:

-5, -3, -3, 1

3 0
2 years ago
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