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notsponge [240]
3 years ago
6

2m-1-5m+4 what is the answer?

Mathematics
1 answer:
KiRa [710]3 years ago
8 0
2m-1-5m+4 

(2m-5m)-1+4

 
   Gather \ like \ terms 

-3m-1+4 

-3m+3

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Solve for x:<br><br> 2.5x - 10.5 = 64 (0.5) ^x
Andreyy89
The answer is x=5. This is because you need to simplify both sides and then isolate the variable.
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3 years ago
6) Find b(10) in the sequence given by b(n) = -5 + 9(n-1).<br> b(10) =?
Sergeeva-Olga [200]

Answer:

76

Step-by-step explanation:

1. Replace n with 10. The equation will look like:

-5+9(10-1)

2. Remember PEMDAS! The first step in solving the equation is solving whatever is in the parenthesis:

-5+9(9)

3. The next step is exponents, but since there aren't exponents in the equation, we move to multiplication and division. The equation should now look like this:

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There's your answer!! Hope this helped :)

6 0
3 years ago
Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen
Jobisdone [24]

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so that the normal vector to S is given by

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

with magnitude

\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}

In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

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\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

where T is the region in the x,y plane over which S is defined. In this case, it's the triangle in the plane z=0 which we can capture with 0\le x\le8 and 0\le y\le\frac{8-x}2, so that we have

\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

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3 years ago
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Firdavs [7]
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Steve and Bart sell cars for KC Complete Auto. Over the past year, they sold 640 cars. Assuming Steve sells 3 times as many as B
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Steve sells 480 cars.

6 0
3 years ago
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