All categories

3 years ago

What is the scientific notation of 1,394,000,000

Mathematics

2 answers:

Ludmilka [50]3 years ago

4 0

$1.394*10^{9}$

White raven [17]3 years ago

3 0

$1\underbrace{394000000}_{\leftarrow9}=1.394\times10^9$

You might be interested in

Solve using standard algorithm 57.03+2.08

Rom4ik [11]

Answer:

59.08

Step-by-step explanation:

6 0

2 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

2 years ago

2a−1/2 please help me :(

Korvikt [17]

Answer:

this is unsolvable i think

Step-by-step explanation:

3 0

2 years ago

the graph shows the speed of a vehicle during the final 50 seconds of a journey at the start of the 50 seconds the speed is k me

Simora [160]

Answer:

a) 27 m/s

b) 30 m/s

c) i) 3

ii) Deceleration

Step-by-step explanation:

The question is not complete, the correct question is given as:

The graph shows information about the speed of a vehicle during the final 50 seconds of a journey. At the start of the 50 seconds the speed is k metres per second. The distance travelled during the 50 seconds is 1.35 kilometres.

(a) Work out the average speed of the vehicle during the 50 seconds

(b) Work out the value of k.

(ii) Describe what this gradient represents

Answer:

The graph is attached. The total time = 50 seconds, total distance = 1.35 km = 1350 m

a) The average speed is the ratio of the total distance traveled to the total time taken to cover this distance. The average speed is given by the formula:

$Average \ speed=\frac{total\ distance}{total\ time}\\ Substituting: \\ Average \ speed=\frac{1350\ m}{50\ s} = 27\ m/s$

b) From the graph, the total distance covered is the area of the graph. The graph is made up of a rectangle and triangle, the area of the graph is equal to the sum of area of rectangle and area of triangle.

$Total \ distance=Total\ area=Area\ of \ rectangle+Area\ of \ triangle\\Total \ distance=(length *breadth)+\frac{1}{2}base*height\\1350 \ m=(40*k)+0.5*10*k\\1350=40k+5k\\45k=1350\\k=1350/45\\k=30\ m/s$ c) i) The gradient in the last 10 seconds is the ratio of change in speed to change in time

$Gradient=\frac{change\ in\ speed }{change\ in\ time}=\frac{0-k}{50-40} \\ bt \ k=30\\Gradient=\frac{0-30}{50-40}= \frac{-30}{10} =-3$

ii) Since the gradient is negative it means it is deceleration. That is in the in the last 10 seconds the vehicle decelerates at a rate of 3 m/s²

6 0

3 years ago

Escribe dos fracciones equivalentes de 12 20

Shalnov [3]

24/40 multiplicado por 2 y 6/10 dividiendo entre 2 las dos son equivalentes

3 0

3 years ago