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3 years ago

−6x+2x−5+2: simplified

Mathematics

1 answer:

Anestetic [448]3 years ago

3 0

Answer:

-4x-3

Step-by-step explanation:

all you do is combine like terms: -6x+2x=-4x and -5+2=-3

so the answer would be -4x-3

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Enter a numerical expression that represents the product of seven squared and forty-six.

Elden [556K]

Iiiiiiuuuuuuuuuuhgfkfgjkvhv

7 0

3 years ago

Evaluate the following expressions using the chip method. SHOW ALL WORK!!!

Mila [183]

Answer:

a. -7
b. -20
c. 7

Step-by-step explanation:

a. -9+2, in this case, it is -7 because you take the bigger number and subtract it by the lower number. If the bigger number is negative your answer will be negative, if the bigger number is positive it will be positive it is just really a basic subtraction problem just add the sign.
b. In multiplication +++=+ ++-=- and a -+-=+ do your problem without thinking about the signs and then add the signs with the formula I showed you.

c. ---=+

Hope this helps :)!

5 0

3 years ago

In 5 years kate will be twice as old as joey. right now kate is 11 years older then joey. right now how old is joey?

natima [27]

K = Kate's age nowJ = Joey's age nowK+5 = Kate's age in 5 yearsJ+5 = Joey's age in 5 years From the first sentenceK+5 = 2(J+5) From the second sentenceK=J+11 So our system of equations isK+5 = 2(J+5)K=J+11 The first equation is equivalent toK+5 = 2J+10or by rearranging termsK-2J=5 The second equation is equivalent toK-J=11 This gives us K-2J=5K-J=11 Using the elimination method, we need to multiply one of the equations by a factor such that we can eliminate one of the variables. This means we want one of the variables to have coefficients which are the same magnitude but opposite signs. We can do this by multiplying the 1st equation by -1. This give us -K+2J=-5 K - J = 11 Now add these two equations together, term by term. This gives usJ=6 So Joey is 6 years old now.

3 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

How many counting numbers less than 20 have an odd amount of factors?

pochemuha

1,3,5,7,9,11,13,14,15,17,19 11 numbers

5 0

4 years ago

Read 2 more answers