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Ket [755]
3 years ago
13

Please give this one a go! I need it ASAP. Thank you for your time

Mathematics
1 answer:
Kisachek [45]3 years ago
4 0

Answer:

Step-by-step explanation:

f(x)=2x+3

f(x)=0

2x+3=0

2x=-3

x=-3/2

x=-1.5

when x=0,y=f(x)=2(0)+3=3

so one point is (0,3)

when x=2,y=f(x)=2(2)+3=4+3=7

so point is (2,7)

Draw a line through  (0,3) and (2,7)

see where it cuts x-axis.

it is the zero of the linear equation.

zero is x=-1.5

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On a unit circle, the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distan
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Check the picture below.

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\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}

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Here is all you need for b)

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