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3 years ago

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After releasing of radioactive material into.The atnosphere from.A nuclear.Powe plant in a country in 1988, the hay in that coun

try was contaminayed by radioactive.Isotope remains, how long did the famers need to.Wait to use this hay

1 answer:

Radda [10]3 years ago

4 0

Answer:

the question is incomplete, so I looked for a similar one:

<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>

iodine's half life (we are given x, we need to find b):

0.5A₀ = A₀eᵇˣ

x = 8 days

we eliminate A₀ from both sides

0.5 = eᵇ⁸

ln 0.5 = ln eᵇ⁸

-0.69315 = b8

b = -0.69315 / 8 = -0.08664

since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):

0.1A₀ = A₀eᵇˣ

0.1 = eᵇˣ

ln 0.1 = bx

where b = -0.08664

x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days

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Let the measure of an angle be 'x'
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3 years ago

I need help with this one.

Mandarinka [93]

Answer:

$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ with T₁ = 5000

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

r = $\frac{1000}{5000}$ = $\frac{200}{1000}$ = $\frac{40}{200}$ = $\frac{1}{5}$

This indicates the sequence is geometric with recursive rule

$T_{n}$ = r $T_{n-1}$

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$T_{n}$ = $\frac{1}{5}$ $T_{n-1}$ , T₁ = 5000

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3 years ago

Which equation represents a proportional relationship? A. y = x + 1 3 y=x+13 B. y = 1 − 1 3 x y=1-13x C. y = 3 x + 1 3 y=3x+13 D

34kurt

Answer:

Option D. $y=\frac{1}{3}x$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

<u><em>Verify each case</em></u>

case A) we have

$y=x+\frac{1}{3}$

Remember that

the line must pass through the origin

so

For x=0, y=0

In this case

For x=0

$y=0+\frac{1}{3}=\frac{1}{3}$

so

The line not passes through the origin

therefore

The equation A not represent a proportional relationship

case B) we have

$y=1-\frac{1}{3}x$

Remember that

the line must pass through the origin

so

For x=0, y=0

In this case

For x=0

$y=1-\frac{1}{3}(0)=1$

so

The line not passes through the origin

therefore

The equation B not represent a proportional relationship

case C) we have

$y=3x+\frac{1}{3}$

Remember that

the line must pass through the origin

so

For x=0, y=0

In this case

For x=0

$y=3(0)+\frac{1}{3}=\frac{1}{3}$

so

The line not passes through the origin

therefore

The equation C not represent a proportional relationship

case D) we have

$y=\frac{1}{3}x$

Remember that

the line must pass through the origin

so

For x=0, y=0

In this case

For x=0

$y=\frac{1}{3}(0)=0$

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The line passes through the origin

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The equation D represent a proportional relationship

3 0

3 years ago

Which of these is true?

Furkat [3]

Answer:

C

Step-by-step explanation:

Tell me if you need or want an explanation

5 0

3 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Step2247 [10]

Answer:

$A = 297\pi$

Step-by-step explanation:

To solve this problem we need to be familiar with the formula for the surface area of a cone:

$A = \pi r(r + \sqrt{h^2+r^2})$

We are given the length of a side and the diameter, to calculate the radius divide the diameter in half:

$r = \frac{d}{2}\\r = \frac{18}{2}\\r = 9 cm$

To calculate the height of the cone, we must use the Pythagorean Theorem:

$C^2 = A^2 + B^2$

We can treat the side length as the hypotenuse $C$ , the radius as the base $A$ , and solve for height $B$ . Set the expression up like this:

$C^2 = A^2 + B^2\\24^2 = 9^2 + B^2\\B^2 = 24^2 - 9^2\\B = \sqrt{24^2 - 9^2}\\B = \sqrt{576 - 81}\\B = \sqrt{495}\\B \approx 22.25$

Now we can plug into our original formula:

$A = \pi r(r + \sqrt{h^2+r^2})\\A = \pi 9(9+\sqrt{\sqrt{495}^2+9^2}\\A = \pi 9(9+\sqrt{495 + 81}\\A = \pi 9(9+\sqrt{576})\\A = \pi 9(9 + 24)\\A = \pi 9(33)\\A = 297\pi$

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2 years ago