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Hi Jenny,

Use desmos graphing calculator and you can check your answers for yourself  

If you are a smart man you can work this out for yourself.  Just draw a rough number line.

200*0,967/0,07 = 2762.8571428571428571

Hi Juriemagic,

If x is negative then f(x) would have to be positive (so the product of them is negative)

likewise

If x is positive then f(x) will have to be negative.

Does that help?

Do you have a specific question that is troubling you?

You can work out the diagonal in terms of s  

and as a separate issue you can work it out in terms of r.

Then you can put them equal to each other to get r in terms of s   OR s in terms of r

The question should be: what is the ratio of the area of the shaded region to the area of both circles?

That is true but it can be written in terms of just s or just r.

C = Capacity of the tank.

1/8C + 20 =3/4C, solve for C

C = 32 Gallons the capacity of the tank.

Consider \((r + s)^2 = r^2 + s^2 + 2rs\).

\(r + s = -\dfrac43\\ rs = 4\)

So

\(\left(-\dfrac43\right)^2 = (r^2 + s^2) + 2\cdot 4\\ r^2 + s^2 = -\dfrac{56}9\)

The midpoint of AB is M(7,-7). If the coordinates of A are (8,-6), whatare the coordinates of B?

Hello Guest!

\(m=\frac{y_A-y_M}{x_A-x_M}=\frac{-6+7}{8-7}=1\)

\(m=1\)

\({\color{blue}y=m(x-x_M)+y_M}=1(x-7)-7\)  (Point-direction equation)

\(x_A-x_M=8-7=1\)

\(x_A-x_M=1\)

\(x_B=x_M-(x_A-x_M)=7-1=6\)

\(x_B=6\)

\(f(6)=6-14=-8\)

\(y_B=-8\)

The coordinates of B are (6, -8)

  !

Find the equation of a line in slope intercept form that is parallel to the line y = -2x - 3 that goesthrough the point (2, 1).

Hello Guest!

Point-directional shape  \(y=m(x-x_1)+y_1\)

\(m=-2\)

\(y=-2(x-2)+1\)

The equation is            \(y=-2x+5\)

  !

You would subtract the two circles' area from the square's area. 

                                              A_square  =  s²  

                                              A_circle     =  πr²  

but there are 2 circles so 

the total area of the circles is                  2πr²  

subtract the area of the circles  

and what's left is the area of the  

shaded region of the square                 s² – 2πr²     and there you have it.  Simpler than you thought it would be. 

_.

Can you think of an example of a question for which you cannot come up with a testable hypothesis?  

What happens to an object when it exceeds the speed of light?  

Albert Einstein's theory says that it's impossible for an object with mass to exceed the speed of light.  Therefore, the question cannot be tested empirically.  

By the way, scientists say that space can exceed the speed of light because it has no mass. 

_.

tangent of x  =  370 / 294  

          tan(x)  =  1.2585034  

          tan^–1 (1.2585034)  =  51.6^o 

_.

we have 3/4-1/8 which is 5/8. 5/8 is 20 gallons. 1/8 is 4 gallons. that means 8/8 is 32 gallons

Hmmm...isn't that what I said? But my answer is clearly wrong, but I can't seem to figure out how or why...

1/   The radius of a small circle is 5

2/   Side of the square is 10

3/   The radius of a bigger circle is   10* sin(45º) = √50

4/   Area of the bigger circle is 50pi  

Hmm...that circle with squares and more circles in it looks like my grandma's coffee table, minus all the tea...

We could just use a ruler, and measure my grandma's coffee table in the matter of a few seconds, but instead, we have decided to spend 10 minutes solving a math problem. Great 👍.

To find the area of the circle, we need to find the radius. And the radius of the big circle is just a simple way to say the diagonal of the square, which can be found using the diameter of the smaller circle. So many circles.

(ain't them pretty?)

Anyways, we first need to find the radius of the small circle. This will be the side lengths of the square, which will help us find the diagonal.

RADIUS OF SMALL CIRCLE:

Ever wonder what the area of the above circle is? Neither did I. 

But, the general formula for a circle's area is:

A = r² pi

Lovely. How does that help us? Well, it just straight out gave away that our radius was 5! (Don't believe me? Try it yourself)

Now, we can go on to the diagonal.

DIAGONAL:

Pythagorean, Pythagorean, Pythagorean. You thought you were done with it once you passed Algebra 1, but it always will haunt you, and your future plans...

In other words, we care looking for:

sqrt(100 + 100) = sqrt(200)

This is 10sqrt(2).

AREA OF BIG CIRCLE:

Finally! What we have gotten what we have been looking for....or at least started to search for it.

The radius of the circle is 5sqrt(2) (which is half of the diagonal)

Plug it into the area formula:

5sqrt(2)² pi = 50pi

There, an answer. Though it still would have been much easier if we just measured my grandma's coffee table...

:)

EDIT: hmmm...this is not possible....but I don't see my mistake...

EDIT 2: thanks to asinus and guest, I was able to detect my mistake!

1/7  +  1/9 = 1 / J

J =3.9375 hours - for both working together.

Solve for x:
3 x^2 + 4 x + 12 = 0

Divide both sides by 3:
x^2 + (4 x)/3 + 4 = 0

Subtract 4 from both sides:
x^2 + (4 x)/3 = -4

Add 4/9 to both sides:
x^2 + (4 x)/3 + 4/9 = -32/9

Write the left hand side as a square:
(x + 2/3)^2 = -32/9

Take the square root of both sides:
x + 2/3 = (4 i sqrt(2))/3 or x + 2/3 = -(4 i sqrt(2))/3

Subtract 2/3 from both sides:
x = (4 i sqrt(2))/3 - 2/3 or x + 2/3 = -(4 i sqrt(2))/3

Subtract 2/3 from both sides:

r = (4 i sqrt(2))/3 - 2/3          or            s= -(4 i sqrt(2))/3 - 2/3

(2/3 (-1 - 2 i sqrt(2)))^2 + (2/3 (-1 + 2 i sqrt(2)))^2 = - 56 / 9

Help out if you can!

Hello Guest!

\(r=\sqrt{\frac{25\pi}{\pi}}\\ r=5\)

\(D=2r\cdot \sqrt{2}\)

\(A=\frac{D^2\pi}{4}=\frac{(2r\cdot \sqrt{2})^2\pi}{4}\)

\(A=50\pi\)   [C]

  !

< The probability is 1/20 >   

That would mean there are five of them.  What are those five numbers, please? 

_.

r^3 - s^3 = 5211/4,

r - s = 3, solve for  r, s

r = -21/2   and   s = -27/2

r = 27/2    and   s = 21/2

Let's call the fraction that it bounces back up "x"  

After the 1st bounce, the ball is  250 • x  high 

&, after the 2nd bounce, the ball is  (250 • x) • x  high 

&, after the 3rd bounce, the ball is [ (250 • x) • x ] • x  high 

All those parentheses aren't necessary, I just included them as a visual

to show that each bounce is x times the height of the previous. 

So.... anothe way of writing the height after the 3rd bounce is 250 • x³  

The height after the 3rd bounce is 16 so let's set the two expressions equal. 

                                                                       250x³  =  16  

                                                                             x³  =  16/250 

Let's change that to a decimal so I can

take the third root with my calculator                     x³  =  0.064 

Take the third root and obtain                                  x  =  0.4 

And, of course, 0.4 = 2/5  so the answer is .... wait for it .....   B. 2/5  

_.

Solve for m:
25 + (m - 5)^2 + m^2 + (m + 5)^2 = 318

Expand out terms of the left hand side:
3 m^2 + 75 = 318

Subtract 75 from both sides:
3 m^2 = 243

Divide both sides by 3:
m^2 = 81

Take the square root of both sides:

 m = 9                                     or     m = -9[discard]