Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with $\sqrt{3}$ $sqrt3$ ?

Question

Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with $\sqrt{3}$ $sqrt3$ ?

I'm having some trouble figuring out how to do this, any chance someone could give an explanation how to solve it?

Thanks

1 Answer

Impact of this question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-25T00:28:58

See below. The answer is $\sqrt{3} ≅ 1.7$ $sqrt(3)~=1.7$ .

Explanation:

[What is linear interpolation?]
Linear interpolation is a method of curve fitting with linear plynomials.
Suppose two points $A (x_{0}, y_{0})$ $A(x_0,y_0)$ and $B (x_{1}, y_{1})$ $B(x_1,y_1)$ are given. If $x_{0} \leq x \leq x_{1}$ $x_0<=x<=x_1$ , the value $y$ $y$ can be approximated by substituting $x$ $x$ to the formula of the line $A B$ $AB$ , that is, $y - y_{0} = \frac{y_{1} - y_{0}}{x_{1} - x_{0}} (x - x_{0})$ $y-y_0=(y_1-y_0)/(x_1-x_0)(x-x_0)$ .
https://en.wikipedia.org/wiki/Linear_interpolation

[Solve this question]
In this question, $\sqrt{1} = 1$ $sqrt(1)=1$ and $\sqrt{4} = 2$ $sqrt(4)=2$ can be used for linear interpolation.

The formula of the line between $A (1, 1)$ $A(1,1)$ and $B (4, 2)$ $B(4,2)$ is as follows.
$y - 1 = \frac{2 - 1}{4 - 1} (x - 1)$ $y-1=(2-1)/(4-1)(x-1)$
$y - 1 = \frac{1}{3} (x - 1)$ $y-1=1/3 (x-1)$
$y = \frac{1}{3} x + \frac{2}{3}$ $y=1/3 x + 2/3$

Substitute $x = 3$ $x=3$ to the formula. The result is $y = \frac{5}{3} = 1.66 \dots ≅ 1.7$ $y=5/3=1.66…~=1.7$

[This is your turn]
Then, how about trying to approximate $\sqrt{3}$ $sqrt(3)$ to the nearest hundredth using interpolation between $C (2.89, 1.7)$ $C(2.89, 1.7)$ and $D (3.24, 1.8)$ $D(3.24,1.8)$ ?
The answer will be $\sqrt{3} ≅ \frac{303}{175} = 1.731 \dots ≅ 1.73$ $sqrt(3)~=303/175=1.731…~=1.73$ .

Science

Math

Humanities

... and beyond

Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with $\sqrt{3}$ $sqrt3$ ?

I'm having some trouble figuring out how to do this, any chance someone could give an explanation how to solve it?

Thanks

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with √3sqrt3?

I'm having some trouble figuring out how to do this, any chance someone could give an explanation how to solve it? Thanks

1 Answer

Explanation:

Related questions

Impact of this question

Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with $\sqrt{3}$ $sqrt3$ ?

I'm having some trouble figuring out how to do this, any chance someone could give an explanation how to solve it?

Thanks