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3 years ago

Please help ASAP!! Thank you in advance!

Mathematics

2 answers:

dexar [7]3 years ago

4 0

Answer:

$\huge\boxed{9) Option \ 2}$

$\huge\boxed{10) \ Option \ 1}$

Step-by-step explanation:

Part 9)

1) (13)(3) = 39

2) (-12)(-3) = 36 ← Correct Option

3) 4(-9) = -36

4) (-18)(2) = -36

Part 10)

= (-3)(-2)(1)(4)

= (6)(1)(4)

= (6)(4)

= 24

$\rule[225]{225}{2}$

Hope this helped!

valkas [14]3 years ago

3 0

Answer:

9.the second one

10.the first one

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3 years ago

Find the equation of a straight line passing throught the point listed and having the given gradient. Express your answer in the

Igoryamba

First let's try to find the equation in this form : y = mx + c

The gradient is given 3 . In a line's equation, x's coefficient represents the line's gradient.

So equation of a line with the gradient of 3, would look like this ;

 $y= 3x + c$

Now a point that the line passes through is given, (1, 2)

This point's x-coordinate is 1 and y-coordinate is 2.

So we'll plug its x-coordinate value in the equation and also y-coordinate value. So we can solve it.

As you know, $x=1$ and $y=2$

$y = 3x + c$

$2\quad =\quad 3\cdot 1+c\\ \\ 2\quad =\quad 3+c\\ \\ 2-3\quad =\quad c\\ \\ -1\quad =\quad c$

We found c = -1

Also in a line's equation, c is constant and it represents the line's y-intercept

So let's build the line's equation.

$m=3$ and $c=-1$

$y= mx + c$

$y= 3x -1$

We found the line's equation in this form, $y= mx + c$

Now let's turn it into this form, $ax + by + c = 0$

$y\quad =\quad 3x-1\\ \\ y-3x\quad =\quad -1\\ \\ y-3x+1\quad =\quad 0\\ \\ -3x+y+1\quad =\quad 0$

Final answers,

$\boxed { y\quad =\quad 3x-1 }$

and

$\boxed { -3x+y+1\quad =\quad 0 }$

I hope this was clear enough :)

5 0

3 years ago

Read 2 more answers

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GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$