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3 years ago

A 33 gram sample of a substance that's used to detect explosives has a k-value of 0.1467.

Mathematics

1 answer:

romanna [79]3 years ago

3 0

Answer:From the attached graphic, Half-life = ln (.5) / k

Half-life =.693147 / 0.1142

= 6.0695884413 days

Step-by-step explanation:The value of "k" should be negative and should have units associated with it.

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A square field has an area of 479ft^2. What is the approximate length of a side of the field? Give your answer to the nearest fo

oksano4ka [1.4K]

For any square the area is given by $A=s^{2}$ where $s$ is the length of one side. (This comes from the general rectangle formula $A=lw$ with $l=w=s$ ).

To solve for $s$ we take the square root of both sides:
$A=s^{2}$
$\sqrt{A}=\sqrt{s^{2}}$
$s=\sqrt{A}$

Now we can substitute known values:
$s=\sqrt{A}$
$s=\sqrt{479}$
$s=21.886068 \approx 22feet$

6 0

3 years ago

What is 60ft/s=__km/h?

vlabodo [156]

Answer:

65.837 km/h to nearest thousandth

Step-by-step explanation:

60 ft/s = 60 * 3600

= 216,000 ft /h

1 foot = 30.48 cms so

216,000 ft/h = 216,000 * 30.48

= 6583680 cms / h

= 6583680 / 100,000

= 65.837 km/h to nearest thousandth

4 0

3 years ago

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

The length of a rectangle is 3 time the width. the perimeter is 65.6 cm. find the width.

Sever21 [200]

The width is 8.2
Perimeter = 2W + 2L (there are four sides of a rectangle, two longer sides, two shorter, W and L)

65.6 = 2W + 2L
L = 3W so substitute 3W in for L 
65.6 = 2W +2 (3W) 
65.6 = 8W 
W = 8.2 cm 
L = 3(8.2) 
L = 24.6cm

5 0

4 years ago

Read 2 more answers

Use the Distributive Property to write an expression equivalent to 7(3x + 9)

ivanzaharov [21]

Answer:

Step-by-step explanation:

5 0

3 years ago