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3 years ago

0.00005026 as a scientific notation

Mathematics

2 answers:

anzhelika [568]3 years ago

8 0

Answer:

5.026 x 10-5

Step-by-step explanation:

Annette [7]3 years ago

3 0

Answer:

5026*10^-5

Step-by-step explanation:

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The amount y (in grams) of the radioactive isotope fermium-253 remaining after t hours is y=a(0.5)t/72, where a is the initial a

almond37 [142]

Answer:

Per hour decay of the isotope is 0.96%.

Step-by-step explanation:

Amount of radioactive element remaining after t hours is represented by

$y=a(0.5)^{\frac{t}{72}}$

where a = initial amount

t = duration of decay (in hours)

Amount remaining after 1 hour will be,

$y=a(0.5)^{\frac{1}{72} }$

y = 0.9904a

So amount of decay in one hour = a - 0.9904a

= 0.0096a gms

Percentage decay every hour = $\frac{\text{Amount of decay}}{\text{Initial amount}}\times 100$

= $\frac{0.0096a}{a}\times 100$

= 0.958 %

≈ 0.96 %

Therefore, per hour decay of the radioactive isotope is 0.96%.

7 0

3 years ago

Which ordered pair is a solution to the inequality?<br><br> Help please!!

Murrr4er [49]

Choice A is the answer which is the point (1,-1)

=========================================

How I got this answer:

Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.

-------------

checking choice A

plug in (x,y) = (1,-1)

$y - 2x \le -3$

$-1 - 2(1) \le -3$

$-3 \le -3$

This is true because -3 is equal to itself. So this is the answer.

-------------

checking choice B

plug in (x,y) = (2,4)

$y - 2x \le -3$

$4 - 2(2) \le -3$

$0 \le -3$

This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.

-------------

checking choice C

plug in (x,y) = (-2,3)

$y - 2x \le -3$

$3 - 2(-2) \le -3$

$7 \le -3$

This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.

-------------

checking choice D

plug in (x,y) = (3,4)

$y - 2x \le -3$

$4 - 2(3) \le -3$

$-2 \le -3$

This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.

4 0

3 years ago

I was walking to the store at the same time as my friend Angela. No matter what, She has always gotten there twice as fast as me

atroni [7]

Let's call your speed "a"

her walking speed is then "2a"

and her friend's scooter speed is 10 a

let's also call the distance you need to walk "d"

so the time (t)you need is:

d/t=a
t=d/a

and her time is:
t= $\frac{d}{2} /2a+ \frac{d}{2} /10a$
(half a distance with the speed of 2a and half a distance (d/2) with the distance of 10 a

now we count (we know that her time is 30 minutes)

30= $\frac{d}{4a}+\frac{d}{20a}$ //multiply by 20a

600a= $5d+d$
600a=6d/divide by 6
100a=d

Now, in order to calculate your time we need to calculate:

t=d/a

but we know how much is d now!
t= $\frac{100a}{a}$ =100

which means that you need 100 minutes, or 1 h 40 minutes!!!

6 0

3 years ago

What is an equation of the line that passes through the points (-2,-5) and (-1,1)

alekssr [168]

Answer:

Slope = 6

Step-by-step explanation:

Slope = $\frac{y^{2}-y^{1} }{x^{2}-x^{1} } =\frac{1-(-5)}{-1-(-2)}=\frac{1+5}{-1+2}=\frac{6}{1}=6$

6 0

2 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago