All categories

1 year ago

What does it mean to determine the two positive integers that each value is between??? I'm confused

Mathematics

1 answer:

KIM [24]1 year ago

7 0

Answer:

11. {7, 8}

12. {8, 9}

13. {8, 9}

14. {4, 5}

Step-by-step explanation:

You want consecutive integers that bracket each of the given irrational roots.

<h3>Square roots</h3>

A square root of an integer will be rational only if the integer is a perfect square. The first few perfect squares are ...

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81

You know these because you know your multiplication tables.

The square root of a number between these perfect squares will lie between the roots of the squares.

50 lies between the squares 49 = 7² and 64 = 8². That means √50 lies between 7 and 8.

Your calculator tells you that √50 ≈ 7.0710678, which is a number that lies between the integers 7 and 8.

72 lies between the squares 64 = 8² and 81 = 9². That means √72 lies between 8 and 9.

65 lies between the squares 64 = 8² and 91 = 9². That means √65 lies between 8 and 9.

23 lies between the squares 16 = 4² and 25 = 5². That means √23 lies between 4 and 5.

<em>Additional comment</em>

The purpose of questions like this appears to be to have you make use of your knowledge of integer perfect squares to guess an approximation of a square root.

A calculator can answer these questions immediately. Of course the two consecutive integers are the integer part of the root, and the next higher integer.

The question is not well-posed. The answer to "integers each value is between" could be 1 < √50 < 50. For integers greater than 1, the square root is always smaller than the integer being rooted. We've seen questions like this enough times that we can guess the intention is for you to identify <em>consecutive</em> integers.

<em>Extra credit</em>

Knowing the integer part of the root and the difference between the number and its next lower perfect square, you can approximate the root as follows:

For integer n = a² +b, the root √n lies between a +b/(2a+1) and a +b/(2a).

For example, 65 = 8²+1, so √65 lies between 8 1/17 and 8 1/16.

You might be interested in

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.