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3 years ago

Need help.

1 answer:

3 0

$f(x)=5x+9\\
f'(x)=5$
The derivative is constant. It's equal 5 for any argument.

$f(x)=\dfrac{6}{x}\\
f'(x)=-\dfrac{6}{x^2}\\
f'(-2)=-\dfrac{6}{(-2)^2}=-\dfrac{6}{4}=-\dfrac{3}{2}$

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Step-by-step explanation:

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3 years ago

Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2) – 6 = 0?

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3 years ago

Read 2 more answers

HELP ASAP PLEASE

yulyashka [42]

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_________________________

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7 0

2 years ago

If f(x) = e^X, and W(f, g) = 3e^x, find g(x).

trasher [3.6K]

Answer:

g(x)=3

Step-by-step explanation:

Let's find the answer.

W(f,g)=3e^x which can be written as:

W(f,g)=(3)*(e^x), notice that:

(e^x)=f(x) so:

W(f,g)=3*f(x), establishing:

W(f,g)=g(x)*f(x) then:

g(x)=3

In conclusion, g(x)=3.

5 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago