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2 years ago

Find the length of side AB. Round your answer to the nearest tenth:

Mathematics

1 answer:

NikAS [45]2 years ago

4 0

Answer:

$\displaystyle 3,1 \approx AB$

Step-by-step explanation:

$\displaystyle \frac{AB}{2} = sec\:50 \hookrightarrow 2sec\:50 = AB \hookrightarrow 3,1114476537... = AB \\ \\ 3,1 \approx AB$

$\displaystyle \frac{2}{AB} = cos\:50 \hookrightarrow ABcos\:50 = 2 \hookrightarrow \frac{2}{cos\:50} = AB \hookrightarrow 3,1114476537... = AB \\ \\ 3,1 \approx AB$

Information on trigonometric ratios

$\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ$

I am joyous to assist you at any time.

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3 years ago

Read 2 more answers

Vertex (0,1) x intercepts is - 1 and 1 what is the equation of the parabola?

MrRa [10]

$\bf y=a(x-{{ h}})^2+{{ k}}\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})$

those are the vertex form of a parabola... so hmmm
the vertex of this one is at 0,1 and intercepts or "solutions" are at -1 and 1, so is opening downwards, notice the picture below

that means, the squared variable is the "x", thus the form is $\bf y=a(x-{{ h}})^2+{{ k}}$

now, we know the vertex is at 0,1, and two x-intercepts of $\pm 1,0$

thus $\bf y=a(x-{{ h}})^2+{{ k}}\qquad 
\begin{cases}
h=0\\
k=1\\
when
\\
x=\pm 1\\
y=0
\end{cases}
\\\\\\
\textit{so.. hmmm let us use the point ... say hmmm 1,0}
\\\\
y=a(x-{{ h}})^2+{{ k}}\implies 0=a(1-0)^2+1$

solve for "a", to see what that coefficient is, then plug it back in the vertex form equation

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3 years ago