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Well...... I try to give good answers  !!!!

OK...that's better, Nickolas !!!

Note that if extended PO  to meet MR, at a point  E .....we would create  two rectangles

The top rectangle MNOE  has a base of 4 units and a height of 2 units...so...its area = 4 * 2 = 8 

The bottom rectangle PERQ has a base of 7 units and a height of (3 - 2) = 1 unit

So..its area = 7 * 1  =  7 

So... the total area = 7 + 8  =   15

Does this make sense ???

Hey.....good pic, Nickolas !!!!

The surface area of the three squares  = 3(2)^2  =  12 units^2

We have two equilateral triangles with  area of 1.5....so...their total surface area = 2(1.5) = 3 units^2

So...total surface area =   12 + 3  =   15 units^2

No we cant

The function is positive on the x intervals where it is above the x axis

Note that it is postive from  x = -infinity to x = -2

And from x = 0  to x = 4.....so

(-inf, -2)    and (0, 4)

Thanks, Lag

Here's a graph : https://www.desmos.com/calculator/yvn9temz1q

The solutions occur at x = -5   and x = 3

See here :  https://web2.0calc.com/questions/need-help-with-a-few-different-types-of-problems-today

Well it says to answer it to the nearest tenth of a minute

Per the last one, just plug these into a graphing calculator and it's easy to see the two intersection points are (-4, 7) and (3, 7). The reason this is the answer is that these two values are the only two values where both equations are equal to each other. Sorry I can't provide an example like CPhill. You can also figure this out by setting both equations equal to each other. \(x^2 + 2x - 8 = 7\). Subtracting 7 from both sides gets us \(x^2 + 2x - 15\). You can figure out the rest by putting this into the quadratic formula.

Hope this helps!

Thanks, Lag....

Here is the graph...https://www.desmos.com/calculator/tarsovokoa

The solutions are as Lag said....!!!

Not sure if you just need the x value, but the two solutions are (-1, 0) and (8, 9). Just plug these two equations into a graphing calculator and it's easy to see the two intersection points

Hope this helps!

No prob !!!

Note, josh....if we drew a horizontal line at   y = 1, we would divide the figure up  into two triangles at the top and a trapezoid at the bottom

The top left triangle has a base of 3 and a height of 2, so its area = (1/2)base * height = (1/2)(3)(2) = 3

The top right triangle has a base of 5 and a height of 1....so the area = (1/2)(5)(1) = 5/2 = 2.5

The trapezoid formed has a height of 4  and two parallel bases of 8 and 11

So...its area = (1/2)(height)(sum of the bases)  = (1/2)(4)(8 + 11) = 2(19) = 38

So....the total area =    3 + 2.5 + 38  =  43.5 [ units^2 ]

Thanks for the help CPhill!  

-8. Just plug this equation into a graphing calculator and it's easy to see

Hope this helps!

So what are we solving for here? I'm assuming that the circle has the area of \(9\pi\) . If so, then we should convert radians to degrees, which gives us 340 degrees. So a 20 degree sector. I think that the sector that you're talking about is the one with 340 degrees. To find the ratio of the sector to circle, we get \(\frac {340}{360} = \frac {17}{18}\).So the sector has area \(\frac {17\cdot 9\pi}{18} = \frac {153\pi}{18}\).

Hope this helps!

Thank you so much!!!!

No pic, Nickolas   .....