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2 years ago

What is the quotient of 144 and 12? *

Mathematics

1 answer:

navik [9.2K]2 years ago

5 0

Answer: 12

Step-by-step explanation:

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Matt built a cage for his snake. The top of the cage was in the shape of a right triangle. One leg of the top was 4x + 2 feet. T

mihalych1998 [28]

We can do S=1/2 of (4x+2)+4 * (3x+1) +4. All we are doing is adding 4 feet. I believe you are able to simplify it to 1/2 of 4x+6 * 3x+5

6 0

2 years ago

78 divided by 190 steps

Goryan [66]

78 x 100 = 7800

How many times does 190 go into 780? 
≈ 4....190 x 4 = 760 

780 - 760 = 20...bring down the extra 0 to make it 200. 

How many times does 190 go into 200? 
≈ 1...subtract 190 from 200 to get a remainder of 10 

190 ÷ 7800 ≈ 41

Hope I Helped You!!! :-)

Have A Good Day!!!

5 0

2 years ago

Solve for x. 0.71= 1/2(9.8) x^2.

Fudgin [204]

Answer: 0.38065

in simpler terms x=± 0.145 =±0.38065

Step-by-step explanation: -490x2+71 = 0

Subtract 71 from both sides of the equation :

-490x2 = -71

Multiply both sides of the equation by (-1) : 490x2 = 71

Divide both sides of the equation by 490:

x2 = 71/490 = 0.145

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:

x = ± √ 71/490

The equation has two real solutions

These solutions are x = ±√ 0.145 = ± 0.38065

5 0

2 years ago

What is the equation of a parabola with a directrix of y=2 and a focus point of 0,-2

KiRa [710]

Hope this helped. :)

Any point, (x0,y0)(x0,y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate:

Distance between the point on the parabola to the focusDistance between the point on the parabola to the directrix

To find the equation of the parabola, equate these two expressions and solve for y0y0 .

Find the equation of the parabola in the example above.

Distance between the point (x0,y0)(x0,y0) and (a,b)(a,b) :

(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(x0−a)2+(y0−b)2

Distance between point (x0,y0)(x0,y0) and the line y=cy=c :

∣∣y0−c∣∣| y0−c |

(Here, the distance between the point and horizontal line is difference of their yy -coordinates.)

Equate the two expressions.

(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=∣∣y0−c∣∣(x0−a)2+(y0−b)2=| y0−c |

Square both sides.

(x0−a)2+(y0−b)2=(y0−c)2(x0−a)2+(y0−b)2=(y0−c)2

Expand the expression in y0y0 on both sides and simplify.

(x0−a)2+b2−c2=2(b−c)y0(x0−a)2+b2−c2=2(b−c)y0

This equation in (x0,y0)(x0,y0) is true for all other values on the parabola and hence we can rewrite with (x,y)(x,y) .

Therefore, the equation of the parabola with focus (a,b)(a,b) and directrix y=cy=c is

(x−a)2+b2−c2=2(b−c)y

3 0

2 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

$\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n$

$\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n$

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

3 years ago