All categories

3 years ago

2 * 10^-4 - 1 * 10^-5 Show your work.

1 answer:

7 0

<span>2 * 10^-4 - 1 * 10^-5
= 10^-4 ( 2 - 1 * 10^-1)
= 10^-4 ( 2 - 0.1)
= 10^-4 (1.9)
= 1.9 * 10^-4

hope it helps</span>

You might be interested in

What is P versus NP, its supposed to be one of the hardest questions in the world

iogann1982 [59]

Answer:

P is the set of problems whose solution times are proportional to polynomials involving N's. ... NP (which stands for nondeterministic polynomial time) is the set of problems whose solutions can be verified in polynomial time. But as far as anyone can tell, many of those problems take exponential time to solve.

Step-by-step explanation:

YW!!!

5 0

3 years ago

When the polynomial is written in standard form, what are the values of the leading coefficient and the constant?

raketka [301]

Answer: The answer is C. The leading coefficient is -3 and the constants is 2.

explanation: I just took the test

3 0

3 years ago

Help please and simplify answer in radical form if needed

deff fn [24]

Answer:

$16\sqrt{3}\:\mathrm{units^2}$

Step-by-step explanation:

The area of an equilateral triangle with side length $s$ is equal to $\frac{\sqrt{3}}{4}s^2$ . Since the perimeter is 24 units and there are three equal sides to an equilateral triangle, each side is $24\div 3=8$ units.

Substituting in this value, we get:

$\frac{\sqrt{3}}{4}8^2=\frac{64\sqrt{3}}{4}=\boxed{16\sqrt{3}}$

6 0

3 years ago

Find the volume of the solid generated when R (shaded region) is revolved about the given line. x=6−3sec y, x=6, y= π 3, and

Dmitrij [34]

Answer:

$V=9\pi\sqrt{3}$

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

$V=\int\limits^b_a {\pi r^{2}} \, dy$

where:

$r=6-(6-3 sec(y))$

$r=3 sec(y)$

a=0

$b=\frac{\pi}{3}$

so the volume becomes:

$V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy$

This can be simplified to:

$V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy$

and the integral can be rewritten like this:

$V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy$

which is a standard integral so we solve it to:

$V=9\pi[tan y]\limits^\frac{\pi}{3}_0$

so we get:

$V=9\pi[tan \frac{\pi}{3} - tan 0]$

which yields:

$V=9\pi\sqrt{3}$ ]

6 0

3 years ago

Which ordered pair represent a function

sweet [91]

Answer:

The first set of ordered pairs is a function, because no two ordered pairs have the same first coordinates with different second coordinates. The second example is not a function, because it contains the ordered pairs (1,2) and (1,5).

Hope this helps you with your question.

3 0

3 years ago