Ask question

Ask question

All categories

Svetllana [295]

3 years ago

11

The half life of a radioactive kind of cadmium is 14 years. How much will be left after 28 years, if you start with 72 grams of

it?

1 answer:

nlexa [21]3 years ago

8 0

Answer:

18 grams

Step-by-step explanation:

You might be interested in

Two plants, plant A and B, are being grown in a science experiment. Plant A is 6 cm tall and growing at a rate of 4 cm/day. Plan

Irina18 [472]

By the second day of growth

6 0

3 years ago

melissa began atop a hill that was 1,350 feet high. she descended 120 feet, took a break and then descended 235 more feet. Which

Anna11 [10]

1500 is the answer I did this a while ago I think not sure

5 0

3 years ago

Read 2 more answers

Aaron bought a bagel and 3 muffins for $7.25. Bea bought a bagel and 2 muffins for $6. How much is a bagel and how much is a muf

Valentin [98]

The muffin is 1.25
The bagel is 3.50
You get this by seeing the first person bought 3 muffins and 1 bagel.
The second vought only 1 muffin less, so subtract the 2 totals and that gives you the difference which is the price of the muffin

3 0

3 years ago

-4x+y=23 and 4x-9y=-15

astra-53 [7]

Answer: x= -6; y= -1

Step-by-step explanation:

-4x + y = 23

4x - 9y = - 15

-4x + y = 23

y = 23 + 4x [Add 4x to both side]

4x - 9(23 + 4x) = - 15

4x - 207 - 36x = - 15

-32x - 207 = -15

-32x = 192 [Add 207 to both side]

x = -6 [Divide -32 from both side]

-4x + y = 23

-4(-6) + y = 23 [Plug in -6 for x]

24 + y = 23 [Subtract 24 from both side]

y = -1

5 0

3 years ago

Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1

nikklg [1K]

Given

$\Sigma_{n=1}^\infty15(-4)^n$

The first 10 partial sums are as follows:

$S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\ \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\ \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\ \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\ \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}$

$S_6=\Sigma_{n=1}^{6}15(-4)^n=\Sigma_{n=1}^{5}15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}$

The rest of the partial sums can be obtained in similar way.

7 0

3 years ago