All categories

3 years ago

Divide (8.6 x 108) by (3.2 x 103). express your answer in scientific notation

1 answer:

7 0

(8.6 x 108) x (3.2 x103

place them in order

(8.6 x 3.2) x (108 x 103)

27.52 x (108 x 103)

27.52 x 11124

That is your answer to this question.

You might be interested in

Find the tangent ratio of angle Θ. Clue: Use the slash symbol ( / ) to represent the fraction bar, and enter the fraction with n

max2010maxim [7]

Answer:

15/8

Step-by-step explanation:

tangent = opposite side over adjacent side.

AB= opposite side 15

BC= hypotenuse 17

CA=adjacent 8

in this case it will be 15/8

7 0

3 years ago

Read 2 more answers

Is Trazillion a number

AlekseyPX

No it is not a real number

7 0

3 years ago

Read 2 more answers

Please help me:( The volume of the above solid is

iren2701 [21]

Answer:

the volume of cylinder = πr²h

h = 5cm

r = 6cm

volume = π × 6² × 5

= π × 6 × 6 × 5

= 180πcm³

3 0

2 years ago

Consider the function below. (If an answer does not exist, enter DNE.) g(x) = 190 + 8x3 + x4 (a) Find the interval of increase.

galben [10]

Answer:

The interval of increase of g(x) is $(-6,+\infty)$ .

Step-by-step explanation:

The interval of increase occurs when first derivative of given function brings positive values. Let be $g(x) = 190 + 8 \cdot x^{3} + x^{4}$ , the first derivative of the function is:

$g'(x) = 24 \cdot x ^{2} + 4\cdot x^{3}$

$g'(x) = 4 \cdot x^{2}\cdot (6+x)$

The following condition must be met to define the interval of increase:

$4\cdot x^{2} \cdot (x+6) > 0$

The first term is always position due to the quadratic form, the second one is a first order polynomial and it is known that positive value is a product of two positive or negative values. Then, the second form must satisfy this:

$x + 6 > 0$

The solution to this inequation is:

$x > - 6$

Now, the solution to this expression in interval notation is: $(-6,+\infty)$

3 0

3 years ago

Pls answer this!! someone!!!

oksano4ka [1.4K]

Answer:

A = bh = (17 km)(11 km) = 187 km^2

3 0

3 years ago