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2 years ago

How many squares would have to be shaded so that of the shape is shaded?

Mathematics

1 answer:

Bogdan [553]2 years ago

5 0

2 because there is 18 squares and if you divide 18 by 2 you get 9 which is the fraction and then you just shade in the 2/9

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current rules for telephone area codes allow the use of digits 2-9 for the first digit and 0-9 for the second and third digits.

lukranit [14]

In this question, there are 10 digit phone number which was divided into 3-3-4 digits. Every first digit of the division should be 2-9 which mean only 8 possibilities. The other digits would be 0-9 which mean 10 possibilities. Then the calculation would be:

8*10*10 * 8*10*10 * 8*10*10*10 = 8^3 * 10*7= 512 * 10^7= 5.12 * 10^9

If you convert 400,000,000 into the scientific form it will become 4* 10^8. Since 5.12*10^9 > 4 * 10^8, then it is clear that the number should be enough

5 0

3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Linear equation 4.5 + 1.5y=24 x-y=4

WARRIOR [948]

4.5x + 1.5y = 24 ⇒ 4.5x + 1.5y = 24
1.0x - 1.0y = 4 ⇒ 4.5x - 4.5y = 18
 6y = 6
 6 6
 y = 1
 4.5x + 1.5(1) = 24
 4.5x + 1.5 = 24
 - 1.5 - 1.5
 4.5x = 22.5
 4.5 4.5
 x = 5
 (x, y) = (5, 1)

8 0

3 years ago

(24 ÷ 6 - 2 + 5 × 3 × (-1 - 3) ÷ 4)^2 How do you solve this? Help me I need this ASAP!!

Bond [772]

Answer:

169

Step-by-step explanation:

PEMDAS

Parenthesis first

Add -1 and -3 which is -4

Divide 24 by 6 which is 4

(4-2+5*3*-4÷4)^2

(4-2-15)^2

(2-15)^2

(-13)^2

Which is 169

6 0

2 years ago

This graph shows the solution to which inequality? A.y>=1/3x-2 B.y<1/3x-2 C.y<=1/3x-2 D.y>1/3x-2

Alina [70]

Answer:

Step-by-step explanation:

3 0

2 years ago