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Jobisdone [24]
3 years ago
8

What are the missing angles?

Mathematics
1 answer:
Komok [63]3 years ago
4 0
Can we get a picture?
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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
kenny6666 [7]

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

  =\frac{-5+7}{4}

  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

3 0
3 years ago
Help me with this question plz <br> Solve: <br> 3x + 9 = 4x
vovikov84 [41]
9 = x

subtract 3x from both sides of the equation and you're left with 9=x
5 0
3 years ago
What kind of snake lover mathematics
kodGreya [7K]

Answer:

me is me

Step-by-step explanation:

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2 years ago
Read 2 more answers
Did I do this right?
zavuch27 [327]

Answer:

3CuCl2 + 2Ni → 3Cu + 2NiCl3

8 0
2 years ago
NEED HELP
solmaris [256]

The solutions to the systems of equations are:

1. (2, 3) (see attachment below). [one solution]

2. no solution

3. (3, 13) [one solution]

<h3>Solution to a System of Equations?</h3>

The solution to a system of equations is the x-value and y-value that will make both equations true. It can be found either using a graph, by elimination method, or substitution method as explained below.

1. Using graph to solve y = 2x - 1 and y = 4x - 5:

The solution is the point where both lines intersect which is: (2, 3) (see attachment below). [one solution]

2. Solving using substitution method:

x = -5y + 4 ---> eqn. 1

3x + 15y = -1 ---> eqn. 2

Substitute x = -5y + 4 into eqn. 2

3(-5y + 4) + 15y = -1

-15y + 12 + 15y = -1

-15y + 15y = -1 - 12

0 = -13 (this shows that there is no solution)

3. Using elimination method:

14x = 2y + 16 ---> eqn. 1

5x = y + 2 ---> eqn. 2

1(14x = 2y + 16)

2(5x = y + 2)

14x = 2y + 16 ----> eqn. 3

10x = 2y + 4 -----> eqn. 4

Subtract

4x = 12

x = 12/4

x = 3

Substitute x = 3 into eqn. 2

5(3) = y + 2

15 = y + 2

15 - 2 = y

13 = y

y = 13

The solution is: (3, 13).

Learn more about the solution of a system of equations on:

brainly.com/question/13729904

#SPJ1

5 0
2 years ago
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