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3 years ago

5

Halpppp and you get some hearts and branlists!

1 answer:

yaroslaw [1]3 years ago

4 0

Answer:

153/20

Step-by-step explanation:

We want to have the mixed fraction turned into a improper fraction to make adding them both easier. We would change 5 2/8 to 42/8 and 2 4/10 to 24/10.

We now need to get them a common denominator. The LCM (least common multiple) of 8 and 10 is 40, so we multiply 42/8 by 5/5 getting us 210/40 and 24/10 by 4/4 to get 96/40.

Finally, we add them then simplify. 96/40 + 210/40 is 306/40. We then simplify that to 153/20.

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3x^+5x+2=0 <br> need work shown plz

Alexxandr [17]

Answer:

${3x}^{2} + 5x + 2 \\ = {3x}^{2} + 3x + 2x + 2 \\ = 3x(x + 1) + 2(x + 1) \\ = (3x +2)(x + 1)$

7 0

2 years ago

Read 2 more answers

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

3 years ago

Find the average rate of change of the function f(x), represented by the graph, over the interval [0, 3]. Calculate the average

Over [174]

Answer:

[f(3) - f(0)]/(3-0)

[f(3) - f(0)]/3

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3 years ago

Can someone plz help me?

earnstyle [38]

Answer:

14.8

Step-by-step explanation:

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2 years ago

Write the number eleven and six hundred fifty-one thousandths in standard form.

sweet-ann [11.9K]

The standard form of thr number is 11.651

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3 years ago