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2 years ago

Real life application of APPROXIMATION

Mathematics

2 answers:

Rus_ich [418]2 years ago

7 0

Answer:

Numbers and calculations are approximated by rounding off. Examples are if there is an answer as 76.89 then it is approximated and considerd as 77. We use approximation in day to day life for the rough calculations of numbers and math.

suter [353]2 years ago

5 0

Answer:

We use approximation when we try to figure out the time it would take to reach a certain place by car.
When you're shopping in the grocery store and trying to stay within a budget, for example, you estimate the cost of the items you put in your cart to keep a running total in your head.

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Carl's property is assessed at $8,500. The property tax rate in his city is 1.35%. What is Carl's property tax?

eduard

Answer is $114.75. This is found by taking 1.35% of 8,500.

6 0

3 years ago

Read 2 more answers

Use the definition of continuity and the properties of limits to show that the function f(x)=x sqrtx/(x-6)^2 is continuous at x

qaws [65]

Answer:

The function is continuous at x = 36

Step-by-step explanation:

From the question we are told that

The function is $f(x) = x * \sqrt{ \frac{x}{ (x-6) ^2 } }$

The point at which continuity is tested is x = 1

Now from the definition of continuity ,

At function is continuous at k if only

$\lim_{x \to k}f(x) = f(k)$

$\lim_{x \to 36}f(x) = \lim_{n \to 36}[x * \sqrt{ \frac{x}{ (x-6) ^2 } }]$

$= 36 * \sqrt{ \frac{36}{ (36-6) ^2 } }$

$= 7.2$

Now

$f(36) = 36 * \sqrt{ \frac{36}{ (36-6) ^2 } }$

$f(36) = 7.2$

So the given function is continuous at x = 36

because

$\lim_{x \to 36}f(x) = f(36)$

7 0

3 years ago

Please help thank you.

sveta [45]

Look at Arc DE

DE = 124°

∠A is 1/2 of Arc DE

124/2 = 62°

B.62 Degrees is your answer

hope this helps

5 0

3 years ago

Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?

Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.

$\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1$