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elena-14-01-66 [18.8K]
3 years ago
12

What is the range of

ac{2}{x-3} -5" alt="f(x)=\frac{2}{x-3} -5" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Ksju [112]3 years ago
7 0

Answer:

Just doing a comment for an extra 5 points sorry but it is what it is

Step-by-step explanation:

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Which figure can serve as a counterexample to the conjecture below?
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3 years ago
URGENT PLEASE
iragen [17]

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=7.2\\ b=6.7\\ c=4.5\\ s=\frac{7.2+6.7+4.5}{2}\\[0.5em] \qquad 9.2 \end{cases} \\\\\\ A\sqrt{9.2(9.2-7.2)(9.2-6.7)(9.2-4.5)}\implies A=\sqrt{9.2(2)(2.5)(4.7)} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A\approx 14.704~\hfill

4 0
3 years ago
The equation T squared = A cubed shows the relationship between a planet's orbital period, T, and the planet's mean distance fro
Dafna11 [192]

Answer:

Ty = √k  * Tx

The orbital period of Y has to be be multiplied by √k .   or . k∧1/2

Step-by-step explanation:

The general equation:

T² = A³

is for the case of planet X

Tₓ² = Aₓ³

In the case of planet Y

Ty² = k * Aₓ³ .     and   Aₓ³ = Tₓ²

By substitution:

Ty² =  k *Tₓ²

Ty = √k  * Tx

3 0
3 years ago
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