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3 years ago

What is 58,413,000,000 in scientific notations

Mathematics

2 answers:

Wittaler [7]3 years ago

7 0

Answer:

$\huge\boxed{58,413,000,000=5.8413\cdot10^{10}}$

Step-by-step explanation:

The scientific notation:

$a\cdot10^k$

where

$1\leq a$

We have the number:

$58,413,000,000$

The decimal point comes after the last zero in that number. We have to move it 10 places to the left.

$5\underbrace{8,412,000,000}_{10}=5\underbrace{.8412000000}_{\leftarrow10}\cdot10^{10}$

We can skip the trailing zeros.

$58,413,000,000=5.8413\cdot10^{10}$

blagie [28]3 years ago

5 0

Answer:

5.8413 × 10^10

Step-by-step explanation:

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What is the distance between the two end points (5,6) (-4,-7) round to the nearth two decimal places

garri49 [273]

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have:

$(5,\ 6)\to x_1=5,\ y_1=6\\(-4,\ -7)\to x_2=-4,\ y_2=-7$

substitute:

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3 years ago

Solve equation 4y+228=353

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To solve for y, we need to get 4y by itself on one side and then simplify for y.

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3 years ago

1. Determine the intervals on which each function is increasing or

motikmotik

Answer:

$\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}$

Step-by-step explanation:

We have the function:

$f(x)=x^3-\frac{11}{2}x^2-4x$

And we want to determine the intervals for which the function is increasing or decreasing.

We will need to first find the critical points of the function. Note that our original function is continuous across the entire x-axis.

To find the critical points, we need to find the first derivative and then solve for the x. So:

$f^\prime(x)=\frac{d}{dx}[x^3-\frac{11}{2}x^2-4x]$

Differentiate:

$f^\prime(x)=(3x^2)-\frac{11}{2}(2x)-(4) \\ f^\prime(x)=3x^2-11x-4$

Set the derivative equal to 0 and solve for x:

$0=3x^2-11x-4$

Factor:

$0=3x^2-12x+x-4 \\ 0=3x(x-4)+(x-4) \\ 0=(3x+1)(x-4)$

Zero Product Property:

$3x+1=0\text{ or } x-4=0 \\ x=-\frac{1}{3}\text{ or } x=4$

Therefore, our critical points are -1/3 and 4.

We can thus sketch the following number line:

<————-(-1/3)——————————————(4)—————>

Now, let’s test values for the three intervals: less than -1/3, between -1/3 and 4, and greater than 4.

For less then -1/3, we can use -1. So, substitute -1 for x for our first derivative and see which we get:

$f^\prime(-1)=3(-1)^2-11(-1)-4=3+11-4=10>0$

The result is positive.

Therefore, for all numbers less than -1/3,f(x) is increasing (since its derivative is positive).

For between -1/3 and 4, we can use 0. Substitute 0 for our derivative:

$f^\prime(0)=3(0)^2-11(0)-4=-4$

The result is negative.

Therefore, for all numbers between -1/3 and 4, f(x) is decreasing (since its derivative is negative).

And finally, for greater than 4, we can use 5:

$f^\prime(5)=3(5)^2-11(5)-4=16>0$

The result is positive.

Therefore, for all numbers greater than 4, f(x) is increasing (since its derivative is positive.

Therefore:

$\text{$f(x)$ is increasing for $x4$}\\ \text{ $f(x)$ is decreasing for $-\frac{1}{3}$

In interval notation:

$\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}$

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3 years ago