The problem 7,291 ÷ 19 is solved below using partial quotients. Identify what partial quotient was used in each step. Then, identify your final quotient.

Seven thousand two hundred ninety one minus five thousand seven hundred equals one thousand five hundred ninety one minus one thousand five hundred twenty equals seventy one minus fifty seven equals fourteen. Arrows pointing to step one, two, and three. You can not make anymore groups of nineteen. What is left becomes your remainder ill give brainliest

Answers

Answer 1

Using the partial quotient method, it is found that:

The partial quotients used in each step are as follows: 300, 80, The final quotient is of: 383.

Partial quotient method

When division is done applying the partial quotient method, consecutive subtractions are done with n groups of the divisor.

Each partial quotient is the number of groups, and the final quotient is the sum of the partial quotients.

In this problem, the divisor is as follows;

19.

The first subtraction is by 5700, which is 5700/19 = 300 groups of 19, hence the first partial quotient is of 300.

The second subtraction is by 1520, which is 1520/19 = 80 groups of 19, hence the second partial quotient is of 80.

The third subtraction is by 57, which is 57/19 = 3 groups of 19, hence the third partial quotient is of 3.

Then the final quotient is given as follows:

300 + 80 + 3 = 383.

Missing Information

The subtractions are given by the image at the end of the answer.

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The Problem 7,291 19 Is Solved Below Using Partial Quotients. Identify What Partial Quotient Was Used

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

Half-Life uses the formula N () = Noe # No = Initial Amount time h = length of half-life. Ifyou start with 65 milligrams of Chromium 51,used to track red blood cells, which has half-life of 27.7 days: How long will it take for 80% ofthe material to decay? t= (27.7) h5 (65) (h3) 4= (27.7) h; (05)

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For the given data it will take 8.69 days for 80% of the material to decay.

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It is given that, If you start with 65 milligrams of Chromium 51, used to track red blood cells, which has a half-life of 27.7 days.

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[tex]N(t) = N_0(1/2)^ {\frac{t}{t_{1/2}} } \\\\ 52 = 65(1/2)^\frac{t}{27.7} \\\\ 52=65\left(\frac{1}{2}\right)^{\frac{t}{27.7}} \\\\ \frac{65\left(\frac{1}{2}\right)^{\frac{t}{27.7}}}{65}=\frac{52}{65} \\\\ \left(\frac{1}{2}\right)^{\frac{t}{27.7}}=\frac{4}{5} \\\\ \frac{t}{27.7}\ln \left(\frac{1}{2}\right)=\ln \left(\frac{4}{5}\right) \\\\ t =-\frac{27.7\ln \left(\frac{4}{5}\right)}{\ln \left(2\right)}[/tex]

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