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2 years ago

fermium-253 has a half-life of three days. a sample contains 400 g. how long will it take for the sample to decay to 100 g?

2 answers:

6 0

This is a relatively easy half life calculation.
After one half-life period, half of the isotope remains.
After two half-life periods, one quarter of the isotope remains.
That being the case, we see this will take 2 half-life periods or 6 days.

11111nata11111 [884]2 years ago

4 0

Answer: 2 half-life periods

and 6 days

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Your school wants to take out an ad in the paper congratulating the basketball team on a successful season, as shown to the rig

Serga [27]

Answer:

1.4in

Step-by-step explanation:

Length of Photo = 4in

Width of Photo = 3in

Unknown:

Value of X = ?

Solution:

Follow these steps:

Area of a rectangle = l x w

Since the photo is a rectangle; area of photo:

Area of photo = 4in x 3in = 12in²

For the area of the ad;

Length of ad = 4 + x

Width of ad = 3 + x

Given that,

the area of the photo = $\frac{1}{2}$ area of ad

12in² = $\frac{1}{2}$ area of ad

Area of ad = 24in²

Area of the ad;

(4 + x) (3 + x) = 24

12 + 4x + 3x + x² = 24

12 + 7x + x² = 24

x² + 7x = 24 - 12

x² + 7x = 12

x² + 7x - 12 = 0

Using the almighty formula where

a = 1, b = 7 and c = -12

x = $\frac{-b (+/-) \sqrt{b^{2} } - 4ac}{2a}$

x = $\frac{-7 + \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}$ or $\frac{-7 - \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}$

x = 1.4 or -8.4

therefore the answer is 1.4in

x is 1.4in

4 0

1 year ago

A freight company has shipping orders for two products. The first product has a unit volume of 10 cu ft, and it weighs 50 lbs. T

Lady_Fox [76]

Answer:

116 units of the first product

380 units of the second product

Step-by-step explanation:

Product 1 has a unit volume of 10 cu ft

Product 2 has a unit volume of 3 cu ft

The truck has 2300 cu ft of space

Product 1 weighs 50 lbs

Product 2 weighs 40 lbs

The truck can carry 21000 lbs

Let X be the units of product 1

Let Y be the units of product 2

The given information can be expressed as:

10X+3Y=2300...(1)

50X+40Y=21000...(2)

Solving the system of equations:

10X+3Y=2300...(1)

10X=2300-3Y

X=(2300-3Y)/10

Substituting X in (2) we have:

50X+40Y=21000

50[(2300-3Y)/10]+40Y=21000

50[(2300/10)-(3Y/10)]+40Y=21000

50[230-(3Y/10)+40Y=21000

11500-(150Y/10)+40Y=21000

11500-15Y+40Y=21000

11500+25Y=21000

25Y=21000-11500

25Y=9500

Y=380

Substituting Y in (1) we have:

10X+3Y=2300...(1)

10X+3(380)=2300

10X+1140=2300

10X=2300-1140

10X=1160

X=116

So 116 units of the first product and 380 units of the second product can be transported in a single shipment with one truck.

8 0

2 years ago

A=<3,0,-1>, find a vector b such that comp(sub a) b=2.

Anna35 [415]

Compab=a.b/|a|
b=<0,1,−2√10>
a.b= 2√10
|a| = √10
a.b/|a|=2√10 / √10=2

5 0

2 years ago

What did the electrician say to his daughter when she came home at 2am

yaroslaw [1]

Answer:

Check Below!!

Step-by-step explanation:

This is basically a cool joke.

When we are talking about electricians, we think about

wires, ohms, resistors, insulator, conductor etc.

Since electrician will say something to her daughter who came home LATE AT NIGHT (2am), the joke needs to go hand-in-hand.

Thus, the electrician said:

"Why are you insulate?"

which would mean, why are you in so late?

HAHA, Good Laugh!

Thanks

4 0

2 years ago

Find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box

Gala2k [10]

Answer:

The dimensions of the box are 12 cm , 12 cm , 12 cm

Step-by-step explanation:

Let x , y and z be the dimensions of box

Volume of box =xyz=1728

$z=\frac{1728}{xy}$

Surface area of box = $2xy+2yz+2xz=2xy+2y(\frac{1728}{xy})+2x(\frac{1728}{xy})$

Let $f(x,y)=2xy+2(\frac{1728}{x})+2(\frac{1728}{y})$

To get minimal surface area

$\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial x}=0$

$2y-2(\frac{1728}{x^2})=0$

$y=\frac{1728}{x^2}$ ----1

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial y}=0$

$2x-2(\frac{1728}{y^2})=0\\x=\frac{1728}{y^2} \\y^2=\frac{1728}{x}$

Using 1

$(\frac{1728}{x^2} )^2=\frac{1728}{x}$

x=0 and $x^3=1728$

Side can never be 0

So, $x^3=1728$

x=12

$y=\frac{1728}{x^2} \\y=\frac{1728}{12^2}$

y=12

$z=\frac{1728}{xy}\\z=\frac{1728}{(12)(12)}$

z=12

The dimensions of the box are 12 cm , 12 cm , 12 cm

5 0

2 years ago