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vredina [299]
3 years ago
12

(x^2+x-17)/(x-4) solve this

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

Answer:

\displaystyle x + 5 + \frac{3}{x - 4}

Step-by-step explanation:

Since the divisor of \displaystyle x - 4is in the form of \displaystyle x - c,we use what is called Synthetic Division. Now, in this formula, −c gives the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

4| 1 1 −17

↓ 4 20

_________

1 5 3 → \displaystyle x + 5 + \frac{3}{x - 4}

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x² + x - 17]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x, the 5 follows right behind it, and bringing up the rear, \displaystyle \frac{3}{x - 4},giving you the quotient of \displaystyle x + 5 + \frac{3}{x - 4}. However, in this case, since you have a remainder of 3, this gets set over the divisor.

I am joyous to assist you anytime.

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Which equation is the inverse of 5y+4 = (x+3)^2 + 1/2
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MA_775_DIABLO [31]

The inequality which represents the missing dimension x is x≥1.6 or x≥8 / 5.

Given that the area is greater than or equal to 8 square feet and image is attached below.

We want to find the missing inequality x in the form of inequality.

The figure is assumed to be a right triangle with Root = x and perpendicular = 10ft.

As we know the area of ​​a triangle is half the product of the base and the height.

First of all, we will find the area of ​​the triangle by substituting the given values ​​we get

Area=(1/2)×Base×height

Area=(1/2)×x×10

Area=5x      ......(1)

Assume that this area is greater than or equal to 8 square feet.

That means Area≥8ft²      ......(2)

Now we will balance equation (1) and equation (2), we get

5x≥8ft²

Furthermore, they we will divide both sides by 5 we get

(5x)/5≥8/5

x≥8/5

x≥1.6

Therefore, the inequality represents the missing size x when the area is larger or equal to 8 square feet is x ≥1.6ft².

Learn more about the dimension from here brainly.com/question/13271352

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4 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 )

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
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