All categories

3 years ago

enya exchanges $200 for euros (€). Suppose the conversion rate is €1 = $1.321. How many euros should Kenya receive?

Mathematics

1 answer:

scoundrel [369]3 years ago

4 0

Answer:

151.40 Euros

Step-by-step explanation:

You might be interested in

Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0, ⇒ dx = 0

y = 9 t, ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$\oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx$

= 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

$= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$ is conservative.

5 0

3 years ago

Solve the simultaneous equations C^2+d^2=5 and 3c+4d=2

Naily [24]

If 3c+4d=2 then c=(2-4d)/3
And if you put that into the other equation
${\frac{2 - 4d}{3}}^{2} + {d}^{2} = 5$
Then you multiply everything by 3, getting

${(2 - 4d)}^{2} + 3 {d}^{2} = 15$
And then

$2 - 4d + \sqrt{3} d = \sqrt{15}$
Leaving
$d( \sqrt{3} - 4) = \sqrt{15} - 2$
I dont have a calc with me right now but you just pot this into the calculator:
$d = \frac{ \sqrt{15} - 2 }{ \sqrt{3} - 4}$
And when you get the number you insert it into the nest eqation to get c

$c = \frac{2 - 4d}{3}$
So basically

$c = \frac{2 - 4 \frac{ \sqrt{15} - 2}{ \sqrt{3} - 4} }{3}$

6 0

3 years ago

Read 2 more answers

Andrew has 8 cassettes. Mary has x cassettes and Jim has twice as many as Andrew. Together they have four times as Mary has. For

kupik [55]

Answer:

mary=x

andrew=8

jim=2*8=16

together=4*mary

total=8+16=24

24/4=x