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Anuta_ua [19.1K]
3 years ago
5

Help me out hereeeeeeeeee

Mathematics
2 answers:
Alja [10]3 years ago
4 0

Answer:

the second answer

Step-by-step explanation:

I hope it helps

frutty [35]3 years ago
4 0

Answer:

1 5/8

Step-by-step explanation:

5/8 * 3 = 15/8 which is the same as 1 7/8

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Easy...will mark brain..plz help :)
Alborosie

Answer: 4-12x

Step-by-step explanation:

1. Multiply the parenthesis by -6

2. Calculate and Reduce

3. Multiply

6 0
3 years ago
Read 2 more answers
Answer i will rate u the brainiest and give u a extra 50 points problem
MrRa [10]

Answer:

man that takes to much time but this is what I got 4,057.6

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3 years ago
George's bank balance last month was) -£350
kvasek [131]

Answer:

mark me as brainliest please

1112

Step-by-step explanation:

1550-350=1200

1200-88=1112

4 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
3 years ago
Is the relationship shown by the data linear? How can you tell if it is or is not? If so, model the data with an equation.
Marat540 [252]

Answer:

linear and y = 2x + 19

Step-by-step explanation:

the x- values increment by 2 and the values of y increment by 4

This indicates a linear relationship exists in the form

y = mx + c ( m is the slope and c the y-intercept )

m = \frac{rise}{run} = \frac{4}{2} = 2

y = 2x + c ← is the partial equation

to find c use any ordered pair from the table

using (- 5, 9 ), then

9 = - 10 + c ⇒ c = 9 + 10 = 19

y = 2x + 19 ← linear equation

As a check

x = - 7 : y = - 14 + 19 = 5 ← correct value of y

x = - 5 : y = - 10 + 19 = 9 ← correct value of y

x = - 3 : y = - 6 + 19 = 13 ← correct value of y

x = - 1 : y = - 2 + 19 = 17 ← correct value of y


4 0
3 years ago
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