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2 years ago

10

Emergency!!

1 answer:

Svetradugi [14.3K]2 years ago

4 0

Answer:

See below

Step-by-step explanation:

<u>Complete the square</u>

$f(x)=x^2-25x\\\\f(x)+(\frac{-25}{2})^2=x^2-25x+(\frac{-25}{2})^2\\\\f(x)+156.25=x^2-25x+156.25\\\\f(x)+156.25=(x-12.5)^2\\\\f(x)=(x-12.5)^2-156.25\\\\f(x)=(x-\frac{25}{2})^2-\frac{625}{156}$

<u>Identify vertex</u>

<u /> $(h,k)\rightarrow(-\frac{25}{2},-\frac{625}{156})$

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(6x² + 36x² + 12) ÷ (x+6)

astra-53 [7]

Answer:

=42x2+1

Step-by-step explanation:

6x2+36x2+12x+6

=42x2+1

8 0

3 years ago

.. pls help me....!!!!! Determine whether the point (-2,4) satisfy y<3x+5,y=3x+5 or y >3x+5..

nasty-shy [4]

Answer:

$y > 3\cdot x + 5$

Step-by-step explanation:

Let $A(x,y) = (-2,4)$ and $y = 3\cdot x + 5$ . Now we evaluate the given function at $x = -2$ :

$y = 3\cdot x + 5$ (1)

$y = 3\cdot (-2)+5$

$y = -6 +5$

$y = -1$

Which means that $y$ is less than the y-component of A. Therefore, the right answer is $y > 3\cdot x + 5$ .

7 0

2 years ago

Wittaler [7]

Answer:

75

Step-by-step explanation:

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3 years ago

What number goes on top?

Artemon [7]

Answer:

33

Step-by-step explanation:

each upper cell is the sum of the two cells under it. For example 4 and 5 make 9 so 16+17=33

7 0

3 years ago

Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tan(x) −

never [62]

Answer:

$f(x,y)=ln secx+cosx siny+C$

Step-by-step explanation:

We are given that DE

$(tanx-sinx siny)dx+cosxcosydy=0$

We have to determine given DE is exact or not.

Compare it with Mdx+Ndy=0

$M=tanx-sinx siny$

$N=cosxcosy$

$\frac{\partial M}{\partial y}=$ $M_y=-sinxcosy$

$\frac{\partial N}{\partial x}=N_x=-sinxcosy$

Therefore, $M_y=N_x$

If DE is exact then $M_y=N_x$

Hence,it is exact.

$M=\frac{\partial f}{\partial x}=tanx-sinxsiny$

Integrate w.r.t x on both sides

$f(x,y)=\int(tanx-sinxsiny)dx$

$f(x,y)=lnsecx+cosxsiny+\phi(y)$ ...(1)

By using the formula

$\int tanxdx=lnsecx+C,\int sinxdx=-cosx+C$

Differentiate partially equation (1) w.r.t y

$\frac{\partial f}{\partial y}=cosxcosy+\phi'(y)$

By using the formula:

$\frac{d(sinx)}{dx}=cosx$

$N=\frac{\partial f}{\partial y}=cosxcosy=cosxcosy+\phi'(y)$

$\phi'(y)=cosxcosy-cosxcosy=0$

$\phi'(y)=0$

Integrate w.r.t y

$\phi(y)=C$

Substitute the value

$f(x,y)=ln secx+cosx siny+C$

7 0

3 years ago