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GarryVolchara [31]
3 years ago
10

There was a sample of 200 milligrams of a radioactive substance to start a study. since then, the sample has decayed by 5.4% eac

h year. let t be the number of years since the start of the study. let y be the mass of the sample in milligrams. write an exponential function showing the relationship between y and t .
Mathematics
1 answer:
abruzzese [7]3 years ago
5 0
For this case we have a function of the form:
 y = A * (b) ^ t
 Where,
 A: initial amount
 b: decrease rate
 t: time
 Substituting values we have:
 y = 200 * (0.946) ^ t
 Answer:
 
An exponential function showing the relationship between and and is:
 
y = 200 * (0.946) ^ t
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In a figure, OB is the radius of a big semicircle and XB is the radius of the small semicircle. Given that OX = 14 cm, Calculate
oee [108]

Answer:

perimeter of the shaded region = 88 +44+28 =160 cm

Step-by-step explanation:

perimeter of shaded region = length AO + arc OB + arc AB

length AO = radius of bigger circle

radius of bigger circle = OX + OB = 2×radius of smaller circle = 2×14 cm = 28 cm

therefore AO = 28 cm

length of arc oB= half of circumference of smaller circle = \pi×14 = 44 cm

length of arc ab = half of circumference of bigger circle  = \pi×28 =\frac{22}{7}×28= 88

therefore perimeter of the shaded region = 88 +44+28 =160 cm

area of the shaded region = half of area of bigger circle - half of area of smaller circle

                                           =\frac{1}{2} \pi 28^{2} -\frac{1}{2} \pi 14^{2}

                                           =\frac{\pi }{2} (28^{2} -14^{2} )

  solving we gen area of shaded region = 924

7 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
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makkiz [27]
45 × 13 = 585

Which is exactly how much he'd need to pay back his sister and buy the bike. So, it would take him 13 weeks.
8 0
3 years ago
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shutvik [7]

Answer:

#16 is linear but idk what 17 is

3 0
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(PLEASE HELP!!!Which number line can be used to find the distance between (-1,2) and (-5,2)?
KIM [24]

Answer:

I think it is the first number line

Step-by-step explanation:

I think it is this one because it is the only one that shows the placement of -1 and -5. I am sorry if the answer is wrong.

6 0
3 years ago
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