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3 years ago

Which triangle on the coordinate grid is a translation of triangle F?

Mathematics

2 answers:

ivolga24 [154]3 years ago

7 0

Triangle G bcs the SL in Translation means SLIDE ;) old trick I was taught

Lesechka [4]3 years ago

4 0

Triangle G it across from it

I know I’m right

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Graph the function. f ( x ) = 1 3 x 2 − 2 x + 8 f(x)= 3 1 x 2 −2x+8

Reika [66]

Answer:

Ok ill get back to you when I figure it out so see you in 5 minutes!

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3 years ago

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300 is 5% of what number

Ber [7]

At first I learned to do these equations as Part÷Whole=Percent÷100
Part(The portion)=300
Whole=X
Percent=5
So now you have
300÷X=5÷100
Set this up as a fraction and cross multiply.
Now you have 30,000=5x
Now divide both sides by 5 and you get x=6,000
6,000 is your answer.

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3 years ago

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Raymond works in an electronics store and gets a 12 percent employee discount. The original cost of a video game system is $175.

slava [35]

Answer:

A. $154.00

Step-by-Step Explanation:

Okay so, the first step is to get 12% and move the decimal to the left twice like this:

.12

Then, multiply 175 by .12 like this:

175 * .12 = 21

Lastly, subtract 21 from 175 like this:

175 - 21 = $154

6 0

3 years ago

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H(x)=-1/3x^2+2x-4 graph the equation

Andreas93 [3]

You should be able to note that this is a parabola which opens downward. Taking the derivative and setting it to -0- will yield the max of the equation, That is

$y=-\frac{1}{3} x^{2} +2x-4$ ⇒ $\frac{dy}{dx} =-\frac{2}{3} x+2=0$ ⇒ $x=3$

4 0

3 years ago

Verify that the function u(x, y, z) = log x^2 + y^2 is a solution of the two dimensional Laplace equation u_xx + u_yy = 0 everyw

Daniel [21]

Answer:

The function $u(x,y,z)=log ( x^{2} +y^{2})$ is indeed a solution of the two dimensional Laplace equation $u_{xx} +u_{yy} =0$ .

The wave equation $u_{tt} =u_{xx}$ is satisfied by the function $u(x,t)=cos(4x)cos(4t)$ but not by the function $u(x,t)=f(x-t)+f(x+1)$ .

Step-by-step explanation:

To verify that the function $u(x,y,z)=log ( x^{2} +y^{2})$ is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.

$u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})$

$u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})$

then we introduce it in the equation $u_{xx} +u_{yy} =0$

we get that $\frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0$

To see if the functions 1) $u(x,t)=cos(4x)cos(4t)$ and 2) $u(x,t)=f(x-t)+f(x+1)$ solve the wave equation we have to calculate the second partial derivative with respect to x and the with respect to t for each function. Then we see if they are equal.

1) $u_{xx}=-16 cos (4x) cos (4t)$

$u_{tt}=-16cos(4x)cos(4t)$

we see for the above expressions that $u_{tt} =u_{xx}$

2) with this function we will have to use the chain rule

If we call $s=x-t$ and $w=x+1$ then we have that

$u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)$

So $\frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}$

because we have $\frac{\partial s}{\partial x} =1$ and $\frac{\partial w}{\partial x} =1$

then $\frac{\partial u}{\partial x} =f'(s)+f'(w)$

⇒ $\frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)$

Regarding the derivatives with respect to time

$\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)$

then $\frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)$

we see that $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }$

$u(x,t)=f(x-t)+f(x+1)$ doesn´t satisfy the wave equation.

4 0

3 years ago