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3 years ago

What is the factorization??

Mathematics

1 answer:

aivan3 [116]3 years ago

7 0

We first find the zeros of the function , the zeros of the function are those at which points the graph of the function crosses or touches the x axis

here we can see in the graph that graph crosses the x axis at two points , 1 and 7

So the zeros of the function are

x=1 and x=7

It means the factor are

(x-1) and (x-7)

So the factorization is given by

(x-1)(x-7)

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3 years ago

Read 2 more answers

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,

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

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3 years ago

circle centered at the origin has a radius of 5 units. The terminal side of an angle, θ, intercepts the circle in Quadrant IV at

astraxan [27]

Answer:

Step-by-step explanation:

x²+y² = 5²

3²+y² = 25

y² = 16

y = ±√16 = ±4

Since the terminal side is in quadrant IV, y = -4

sinθ = -4/5

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3 years ago

I am very confused lol... So I'm doing a worksheet for Algebra 1 (highschool) and we're doing division and multiplication with f

evablogger [386]

OK first change 3 3/4 to an improper fraction and that will give you 15/4. so you got that part right. Next, change the division sign to multiplication. then flip the 1/2 to 2/1. Then multiply 15 times 2 and that will give you 30. Then multiply 4 times 1 and that will give you 4. then you will have an answer of 30/4. Finally, since you a have a big number over a small number you divide 30 by 4 and that will give you 7 1/2. So it turns out that your little cheat sheet is wrong because when you check your answer you should go on Math-way to check it. Also, a little background about me is that I am in 8th grade and I am taking algebra 1 too.

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3 years ago