QuestionFind the greatest integer less than $\sqrt{286}$ and $\sqrt[3]{70}$ .

Studdy Solution

STEP 1

Assumptions1. We are looking for the greatest integer that is less than the given surds. . We are not allowed to use a calculator.
3. We are dealing with real numbers.

STEP 2

Let's start with the first surd, $\sqrt{286}$ . We know that if $n^2 <286 < (n+1)^2$ , then the greatest integer less than $\sqrt{286}$ is $n$ .

STEP 3

We need to find such $n$ that $n^2 <286$ . We can start by trying out some perfect squares. We know that $16^2 =256$ and $17^2 =289$ , so $16^2 <286 <17^2$ .

STEP 4

Therefore, the greatest integer that is less than $\sqrt{286}$ is16.

STEP 5

Now, let's move on to the second surd, $\sqrt[3]{70}$ . We are looking for an integer $n$ such that $n^3 <70 < (n+1)^3$ .

STEP 6

We know that $4^3 =64$ and $5^3 =125$ , so $4^3 <70 <5^3$ .

STEP 7

Therefore, the greatest integer that is less than $\sqrt[3]{70}$ is4.
The greatest integer less than $\sqrt{286}$ is16 and the greatest integer less than $\sqrt[3]{70}$ is4.

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QuestionFind the greatest integer less than $\sqrt{286}$ and $\sqrt[3]{70}$ .

Studdy Solution

STEP 1

Assumptions1. We are looking for the greatest integer that is less than the given surds. . We are not allowed to use a calculator.
3. We are dealing with real numbers.

STEP 2

Let's start with the first surd, $\sqrt{286}$ . We know that if $n^2 <286 < (n+1)^2$ , then the greatest integer less than $\sqrt{286}$ is $n$ .

STEP 3

We need to find such $n$ that $n^2 <286$ . We can start by trying out some perfect squares. We know that $16^2 =256$ and $17^2 =289$ , so $16^2 <286 <17^2$ .

STEP 4

Therefore, the greatest integer that is less than $\sqrt{286}$ is16.

STEP 5

Now, let's move on to the second surd, $\sqrt[3]{70}$ . We are looking for an integer $n$ such that $n^3 <70 < (n+1)^3$ .

STEP 6

We know that $4^3 =64$ and $5^3 =125$ , so $4^3 <70 <5^3$ .

STEP 7

Therefore, the greatest integer that is less than $\sqrt[3]{70}$ is4.
The greatest integer less than $\sqrt{286}$ is16 and the greatest integer less than $\sqrt[3]{70}$ is4.

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QuestionFind the greatest integer less than 286\sqrt{286}286​ and 703\sqrt[3]{70}370​.

Studdy Solution

STEP 1

Assumptions1. We are looking for the greatest integer that is less than the given surds. . We are not allowed to use a calculator.3. We are dealing with real numbers.

STEP 2

Let's start with the first surd, 286\sqrt{286}286​. We know that if n2<286<(n+1)2n^2 <286 < (n+1)^2n2<286<(n+1)2, then the greatest integer less than 286\sqrt{286}286​ is nnn.

STEP 3

We need to find such nnn that n2<286n^2 <286n2<286. We can start by trying out some perfect squares. We know that 162=25616^2 =256162=256 and 172=28917^2 =289172=289, so 162<286<17216^2 <286 <17^2162<286<172.

STEP 4

Therefore, the greatest integer that is less than 286\sqrt{286}286​ is16.

STEP 5

Now, let's move on to the second surd, 703\sqrt[3]{70}370​. We are looking for an integer nnn such that n3<70<(n+1)3n^3 <70 < (n+1)^3n3<70<(n+1)3.

STEP 6

We know that 43=644^3 =6443=64 and 53=1255^3 =12553=125, so 43<70<534^3 <70 <5^343<70<53.

STEP 7

Therefore, the greatest integer that is less than 703\sqrt[3]{70}370​ is4.The greatest integer less than 286\sqrt{286}286​ is16 and the greatest integer less than 703\sqrt[3]{70}370​ is4.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the greatest integer less than $\sqrt{286}$ and $\sqrt[3]{70}$ .

Assumptions1. We are looking for the greatest integer that is less than the given surds. . We are not allowed to use a calculator.
3. We are dealing with real numbers.

Let's start with the first surd, $\sqrt{286}$ . We know that if $n^2 <286 < (n+1)^2$ , then the greatest integer less than $\sqrt{286}$ is $n$ .

We need to find such $n$ that $n^2 <286$ . We can start by trying out some perfect squares. We know that $16^2 =256$ and $17^2 =289$ , so $16^2 <286 <17^2$ .

Therefore, the greatest integer that is less than $\sqrt{286}$ is16.

Now, let's move on to the second surd, $\sqrt[3]{70}$ . We are looking for an integer $n$ such that $n^3 <70 < (n+1)^3$ .

We know that $4^3 =64$ and $5^3 =125$ , so $4^3 <70 <5^3$ .

Therefore, the greatest integer that is less than $\sqrt[3]{70}$ is4.
The greatest integer less than $\sqrt{286}$ is16 and the greatest integer less than $\sqrt[3]{70}$ is4.