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Zielflug [23.3K]
3 years ago
12

Use the graph to find the solution of the system of equations {y = x^2 + 6x + 8 {y = x + 4

Mathematics
1 answer:
Phantasy [73]3 years ago
8 0
Essentially when people ask you find the solution to system of equation, there asking at what x value do these to graphs intersect. The easiest way to do this is to get a graphing calculator, or desmos and type in the equation and find where they intersect. Heck, even the question says to solve it with a graph, but I'll demonstrate it algebraically.

One way you can do this is set the equation equal to each other. This is because you want to know at what x-value has the same y-value. So we get:

x^2 + 6x + 8 = x + 4

We can then combine like terms, or move everything to one side. So we get:

x^2 + 5x + 4 = 0.

Then we can use the quadratic formula to solve for x.

x=(-5 +/- sqrt(5^2 - 4(1)(4)))/(2(1)

This simplifies into:

(-5 +/- 3)/2

Finally we add and subtract:

(-5 + 3)/2 = x = -1
(-5 - 3)/2 = x = -4

And our solution is x = -1, x = -4
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\huge\mathfrak{\underline{answer:}}

\large\bf{\angle ACD = 105°}

__________________________________________

\large\bf{\underline{Here:}}

  • BCD is an isosceles right triangle , right angled at D
  • ABC is an equilateral triangle

\large\bf{\underline{To\:find:}}

  • ∠ ACD

\large\bf{In\: triangle\:ABC:}

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

\large\bf{\underline{So}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = \frac{180}{3}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle ACB = 60°}

\large\bf{In\: triangle\:BDC}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle D = 90°}

\bf{⟼\angle DBC =\angle DCB }(Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB + 90° = 180°}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼2\angle DCB = 180 -90}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = \frac{90}{2}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟼\angle DCB = 45°}

\large\bf{\underline{Therefore,}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = \angle ACB + \angle DCB}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\bf{⟹\angle ACD = 60+45}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\angle ACD = 105°}

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