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3 years ago

If a number is divisible by four then it is divisible by eight 12 16 24 48

Mathematics

2 answers:

Basile [38]3 years ago

8 0

No because 12 is divisible by 4 but not by 8

Assoli18 [71]3 years ago

3 0

It’s 24 i hope this helps

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1. 52

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3 years ago

at an amusement park, 60% of the people were adults, and the rest were children. there were 720 adults. how many people were at

PIT_PIT [208]

There was 288 children at the amusement park.

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3 years ago

Please help ASAP<br> Algebra 2<br> (the answer is a x and y equation)

Ratling [72]

Answer:

(2, 8)

Step-by-step explanation:

A hole occurs when the denominator is equal to 0. In this case, the denominator equals zero when x = 2.

Next, we must find the value of y by factoring the numerator and simplifying: $\frac{x^{2} +4x-12}{x-2} = \frac{(x-2)(x+6)}{x-2} = x+6$

Then plug in 2 for x:

2 + 6 = 8.

Therefore, there is a hole at (2, 8)

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3 years ago

Use the surface integral in Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus

Tom [10]

Stokes' theorem says the integral of the curl of $\vec F$ over a surface $S$ with boundary $C$ is equal to the integral of $\vec F$ along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

$\displaystyle\iint_S\nabla\times\vec F\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r$

We have

$\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=4\,\vec k$

Parameterize the ellipse $S$ by

$\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k$

Then the flux of the curl is

$\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}$

4 0

3 years ago

Please help!

FrozenT [24]

Points are collinear if they lie on the same line.

First find the equation of the line that passes through the points B and C.
$B(4, -3) \\
x_1=4 \\ y_1=-3 \\ \\
C(-4,3) \\
x_2=-4 \\ y_2=3 \\ \\
m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(-3)}{-4-4}=\frac{3+3}{-8}=\frac{6}{-8}=-\frac{3}{4} \\ \\
y=-\frac{3}{4}x+b \\
(-4,3) \\
3=-\frac{3}{4} \times (-4)+b \\
3=3+b \\
b=0 \\ \\
y=-\frac{3}{4}x$

The points lie on the line y=(-3/4)x.
Now plug the coordinates of the given points into the equation and check if they satisfy the equation.

$(0,0) \\
x=0 \\ y=0 \\ \Downarrow \\
0 \stackrel{?}{=} -\frac{3}{4} \times 0 \\
0 \stackrel{?}{=} 0 \\
0=0 \\
\hbox{the point lies on the line} \\ \\
(1,1) \\
x=1 \\ y=1 \\ \Downarrow \\
1 \stackrel{?}{=} -\frac{3}{4} \times 1 \\
1 \stackrel{?}{=} -\frac{3}{4} \\
1 \not= -\frac{3}{4} \\
\hbox{the point doesn't lie on the line}$

$
(1,-5) \\
x=1 \\ y=-5 \\ \Downarrow \\ -5 \stackrel{?}{=} -\frac{3}{4} \times 1 \\
-5 \stackrel{?}{=} -\frac{3}{4} \\
-5 \not= -\frac{3}{4} \\
\hbox{the point doesn't lie on the line} \\ \\
(6,-8) \\
x=6 \\ y=-8 \\ \Downarrow \\
-8 \stackrel{?}{=} -\frac{3}{4} \times 6 \\
-8 \stackrel{?}{=} -\frac{9}{2} \\
-8 \not= -\frac{9}{2} \\
\hbox{the point doesn't lie on the line}$

The answer is A.

7 0

3 years ago

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