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2 years ago

What is the zero of this function x^2 +3x-4

Mathematics

2 answers:

dexar [7]2 years ago

8 0

-4. This is the answer because all you do is plug zero in for x

Semmy [17]2 years ago

5 0

0^2 + 3(0) -4 = -4 so you plug in 0 for the x's

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How do I do this -9 - 16m = 7? What are the steps?

nordsb [41]

m = -1

Step-by-step explanation:

-16m = 7 + 9

-16m = 16

m = 16 ÷ -16

m = -1

4 0

2 years ago

How do I find the displacement? Help frl

liubo4ka [24]

Answer:

distance traveled: 30 m
displacement: 21.4 m

Step-by-step explanation:

You want the distance traveled and the displacement after walking 17 m south and 13 m east.

<h3>Distance</h3>

The distance traveled is the sum of the lengths of each leg of the trip:

17 m + 13 m = 30 m

You have traveled a distance of 30 m.

<h3>Displacement</h3>

The displacement is the distance from your final position to your starting position. If you draw a diagram of the journey, you see the displacement is the hypotenuse of a right triangle with legs 17 m and 13 m. The Pythagorean theorem can help you find this length:

h = √(a² +b²)

h = √(17² +13²) = √(289 +169) = √458 ≈ 21.401

At the end of your walking, you are 21.4 m from where you started.

5 0

1 year 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

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