All categories

3 years ago

How do I simply this, thanks!

Mathematics

1 answer:

stepan [7]3 years ago

7 0

(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4

You might be interested in

Help please<br> I will give brainliest

Mashutka [201]

tan (<Q)=11/8
m(<Q)=tan^-1(11/8)=53.96°
tan(<P)=8/11
m(<P)=tan^-1(8/11)=36.02°
QP=Square root of (8)^2+(11)^2=square root 185
the answer of the second question is 53°

8 0

3 years ago

When a ferry boat is on open water, it travels at a constant speed. A ferry boat traveled 4 miles in 10 minutes. Complete the eq

ozzi

Answer:

The equation is: $y=0.4x$

The relationship between the time in minutes, t, and the distance traveled in miles, d is a straight line, as shown in attached graph.

Step-by-step explanation:

As boat travels at constant speed.

So, if ferry boat travels 4 miles in 10 minutes. It means it would complete 24 miles in one hour.

And in 1 minute, 0.4 miles will be completed.

So, the equation becomes

$y=0.4x$

As the slope intercept form is $y=mx+c$

So, $m=0.4$ is the slope

The relationship between the time in minutes, t, and the distance traveled in miles, d is a straight line i.e. linear.

The graph is also attached which indicates that relationship between the time in minutes, t, and the distance traveled in miles, d is a straight line i.e. linear.

Keywords: speed, time, graph

Learn more about speed and time from brainly.com/question/1969153

#learnwithBrainly

4 0

3 years ago

Read 2 more answers

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$