What is the next number in this pattern?<br><br> 0, 3, 15, 63

Which function represents transforming ƒ(x) = 3x with a reflection over the x-axis and a vertical shift of 4 units?

Oxana [17]

Answer:

D. $h(x)=-3x+4$

Step-by-step explanation:

we are given

$f(x)=3x$

Firstly, it is reflected over x-axis

we know that for reflection about x-axis , we can replace y as -y

we get

$g(x)=-f(x)$

$g(x)=-3x$

now, it is vertical shift up by 4 units

so, we can add 4 to y-value

we get

$h(x)=g(x)+4$

$h(x)=-3x+4$

4 0

3 years ago

The circle below is centered at (2, 3) and has a radius of 4. What is its

OLEGan [10]

C. Is try correct answer
See example in the attached picture

6 0

3 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

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

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

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

3 0

3 years ago

What is the numerical coefficient in the polynomial 4y+ 6?

GREYUIT [131]

Answer:

4y

Step-by-step explanation:

A coefficient is a number with a variable and no sign in between them.

4 0

2 years ago

What is the value of 13 + 5 x (8 - 6)^2

Nezavi [6.7K]

Answer:

33

Step-by-step explanation:

(8-6)= 2

2^2=4

4 × 5 = 20

20 + 13 = 33

8 0

2 years ago

What is the next number in this pattern? 0, 3, 15, 63