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3 years ago

Construct the line perpendicular to ⇒NO at point P.

Mathematics

2 answers:

suter [353]3 years ago

7 0

The 3rd one is correct, you would put a compass on Point P and draw a small arc on lIne NO on each side of Point P, then place the compass on the line at those points and draw arcs above and below P then connect those points with a ruler.

shepuryov [24]3 years ago

4 0

Answer:

The third one.

Step-by-step explanation:

Basic style of how you construct a perpendicular bisector.

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Help ASAP please and thanks

Firlakuza [10]

The first one includes both

4 0

2 years ago

Read 2 more answers

|0.7x+5|>6.7<br><br>Help plz

Pavlova-9 [17]

Answer:

$\large\boxed{x\dfrac{17}{7}}\\\downarrow\\\boxed{x\in\left(-\infty,\ -\dfrac{117}{7}\right)\ \cup\ \left(\dfrac{17}{7};\ \infty\right)}$

Step-by-step explanation:

$|0.7x+5|>6.7\iff0.7x+5>6.7\ or\ 0.7x+56.7\qquad\text{subtract 5 from both sides}\\0.7x>1.7\qquad\text{divide both sides by 0.7}\\x>\dfrac{1.7}{0.7}\to x>\dfrac{17}{7}\\\\(2)\\0.7x+5$

5 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$