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3 years ago

Solve y/3 -4 = 2 , what does Y equal?

Mathematics

1 answer:

Andreas93 [3]3 years ago

4 0

Answer:

y = 18

Step-by-step explanation:

y/3 - 4 = 2

y/3 = 6

y = 18

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What is the formal definition of a <br>square ?

AfilCa [17]

A shape with 4 equal sides.

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3 years ago

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Y = −(x + 4)2 − 7 vertex

ICE Princess25 [194]

Answer:

The vertex (h,k) is (-4,-7).

Step-by-step explanation:

I assume you are looking for the vertex $y=-4(x+4)^2-7$ .

The vertex form of a quadratic is $y=a(x-h)^2+k$ where the vertex is (h,k) and a tells us if the parabola is open down (if a<0) or up (if a>0). a also tells us if it is stretched or compressed.

Anyways if you compare $y=-4(x+4)^2-7$ to $y=a(x-h)^2+k$ , you should see that $a=-4,h=-4,k=-7$ .

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3 years ago

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The chess club members bought 6 tickets to a tournament for $15. How much would they have paid if all 9 members wanted to go?

olga nikolaevna [1]

Answer:$22.50

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4 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

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