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2 years ago

Millions for defense, but not one cent for trubute" was in reference to what?

Mathematics

1 answer:

Crazy boy [7]2 years ago

7 0

The XYZ affairs
and it what said by Robert Goodloe Harper

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YALL NEVER ANSWER AH POOP plz help

lakkis [162]

Answer:

The given point A (6,13) lies on the equation. True

The given point B(21,33) lies on the equation. True

The given point C (99, 137) lies on the equation. True

Step-by-step explanation:

Here, the given equation is : $y =\frac{4}{3} x + 5$

Now,check the given equation for the given points.

1) A (6,13)

Substitute x = 6 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (6) + 5 = 4(2) + 5 = 8 + 5 = 13$

⇒ y= 13

Hence, the given point A (6,13) lies on the equation.

2) B (21,33)

Substitute x = 21 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (21) + 5 = 4(7) + 5 = 28 + 5 = 33$

⇒ y = 33

Hence, the given point B(21,33) lies on the equation.

3) C (99, 137)

Substitute x = 99 in the given equation $y =\frac{4}{3} x + 5$

$y =\frac{4}{3} (99) + 5 = 4(33) + 5 = 132 + 5 = 137$

⇒ y = 137

Hence, the given point C (99, 137) lies on the equation.

7 0

2 years ago

Read 2 more answers

What is 0.275 0.20 0.572 and 0.725 greatest to least

Andru [333]

.725 , .572 , .275, .2
You look at the first decimal place to see which is the greatest. 7 is greater than 5 and 2 so it's first. Then 5 is greater than 2 so it is next. For the last two numbers since 2 is the same value you look at the next decimal place. Since 7 is greater than 0 it comes next!

6 0

2 years ago

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

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

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />