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2 years ago

6.4=0.8m what does m equal

Mathematics

1 answer:

mestny [16]2 years ago

6 0

Answer:

0.8

Step-by-step explanation:

8x8=64 but put the decimal point back in 0.8x0.8=6.4

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Perform the indicated operation.

GarryVolchara [31]

Answer:

a= 4

b= -10

c= -7

d= -1

Step-by-step explanation:

3 0

2 years ago

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Given the two expressions shown below:

Alchen [17]

A is irrational but b is rational. This is because for b you can work out both square root of 64 (8) and square root of 4 (2). But for a, the square root of 5 is an irrational number :)

5 0

3 years ago

Four numbers from the set 0 to 9, including 0 and 9, are used to create a personal identification number. Once a number is used,

dolphi86 [110]

Answer:

1. 5,040

2. 504

3. 0.1

Step-by-step explanation:

just took the assignment

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2 years ago

Plz guys help me In 2 please

Doss [256]

Answer:

Step-by-step explanation:

OMG!!!!! why don't you just SAY YOU WANT a BLEEPING REFLECTION!!!

I had to search through a slew of your questions to get clues as to what to do.

To reflect over the x axis, just negate the y term and leave the x term as it is.

M becomes (2, -1)

N becomes (-3, -1)

P becomes (-1, -4)

7 0

2 years ago

The core melt- down and explosions at the nuclear reactor in Chernobyl in 1986 released large amounts of strontium-90, which dec

elena-14-01-66 [18.8K]

If $S(t)$ is the amount of strontium-90 present in the area in year $t$ , and it decays at a rate of 2.5% per year, then

$S(t+1)=(1-0.025)S(t)=0.975S(t)$

Let $S(0)=s$ be the starting amount immediately after the nuclear reactor explodes. Then

$S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^{t+1}S(0)$

or simply

$S(t)=0.975^ts$

So that after 50 years, the amount of strontium-90 that remains is approximately

$S(50)=0.975^{50}s\approx0.282s$

or about 28% of the original amount.

We can confirm this another way; recall the exponential decay formula,

$S(t)=se^{kt}$

where $t$ is measured in years. We're told that 2.5% of the starting amount $s$ decays after 1 year, so that

$0.975s=se^k\implies k=\ln0.975$

Then after 50 years, we have

$S(50)=se^{50k}\approx0.282s$

5 0

2 years ago