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3 years ago

TEST ITS TIMED

Mathematics

1 answer:

Aliun [14]3 years ago

5 0

Answer:

Angle 6, Angle 8, Angle 2, and Angle 4 are supplementary to Angle 7.

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Greg goes fishing every day for a week. On the first day he catches seven fish and each

Oliga [24]

so first day and so on

7, 10, 13,....

as you can see it's an arithmetic progression

so sum for nth term= n/2 { 2a + (n-1) d}

it's the sum of the 7th term

7/2 { 7 ×2 + ( 7-1) 3}

7/2 × 32

7× 16

112 fishes

6 0

2 years ago

Read 2 more answers

Whats the radius for the circle? x^2+2x+y^2+4y-6=0

MAXImum [283]

Answer:

r = $\sqrt{11}$

Step-by-step explanation:

So we need to complete the square for both parts of the equation

First though we can add the 6 to the other side so we have x² + 2x + y² + 4y = 6

So first we can complete the square for x² + 2x

To do so we need to use $(\frac{b}{2} )^{2}$ to figure out the number we need to add to both sides

In this case our b is 2, so substituting this in we get $(\frac{2}{2} )^{2} =(1)^{2} =1$

Here we add 1 to both sides and now we have x² + 2x + 1 + y² + 4y = 6 + 1

Now we can follow the same steps to complete the square for y² + 4y

Here our b is 4, so substituting this in we get $(\frac{4}{2} )^{2}=(2)^{2} =4$

Now we add 4 to both sides and now we have x² + 2x + 1 + y² + 4y + 4 = 6 + 1 + 4

Now condensing everything we have (x + 1)² + (y + 2)² = 11

The formula for a circle is (x - h)² + (y - k)² = r²

In our equation we have r² = 11

To find the radius we need to take the square root of both sides $\sqrt{r^{2}} =\sqrt{11}$ to get r = $\sqrt{11}$

5 0

2 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$