All Answers 358083 Answers

1)

When the question asks to solve for h in the equation of \(V=2\pi r^2h\), it means get h, alone, on one side of the equation.  This only requires one step, in this case:

The appropriate answer choice, therefore, is D

2) If the domain is only {2, 4, 7}, just plug them in the original equation of \(f(x)=x^2-5\).

Now, find the answer choice that has all three of these in its domain. That would be D again.

Maxwong has the correct idea, but I think missed the criterion that states that the triangle must be acute. With this restriction, there are not as many possibilities. 

Have you ever heard of the Pythagorean Theorem? You probably have. But have you heard of the Pythagorean Inequalities Theorem? Maybe you have and maybe you haven't. I think my textbook summarizes the concept nicely and in a way that is fairly straightforward:
 

Since we can orient the triangle any way we want ...let  A  =(0,0) , B  = (0, b)  and C  = (c, 0)

Let  D  =   ( c/2 , b/2)    ...so  AD  =  sqrt [ (c/2 - 0)^2   +  (b /2 - 0) ^2  ]  =  sqrt [ ( b^2 + c^2 ) / 4 ]   = 

sqrt [ b^2 + c^2 ] / 2

AB  = b

And 

AC  = c

So

AD  <  [ AB  + AC] / 2

sqrt [ b^2 + c^2  ] / 2  <  [ b + c ] / 2

sqrt [ b^2 + c^2 ] < b + c        square both sides

b^2 + c^2  <  b^2  + 2bc + c^2

0 <  2bc       which is true  since  b, c   are both > 0 

In the second question, Cphill interpreted it correctly, but if you meant \(R(x)=\frac{3x-3}{x^2-4}\) or \(R(x)=3x-\frac{3}{x^2-4}\), the answer is different. Let's think about what we have to do to figure out the vertical asymptote.

In both cases, we have rational fractions. Of course, we can't have a denominator of 0. This means that if we plug in a value for x that results in a denominator that equals 0, then it is officially outside of the domain. Let's figure out when x^2-4=0.

This means that the vertical asymptote is at \(x=\pm2\), which corresponds to the answer choice of C.

And 78.54%  rounded to the nearest inetger =  79%

I have already answered this question. It was posted. Then, it disappeared into the stratosphere. That's strange.

In any case, usually, this expression would be in its simplified form already. However, in this case, the radicands (numbers inside of the radical) happen to be perfect squares, so this expression can be simplified further:

Ratio of smaller circle's circumference to larger circle's circumference  =

   [  2 pi * (1/2)R ] / [ 2 pi * R ]   =

[1/2 R ]  / [ R ]     =    1 : 2   =   x : y

No, the expression evaluates to 39:

I'll help you simplfy this expression \(\left(\frac{1}{2}\right)^2-6\left(2-\frac{2}{3}\right)\):

Therefore, \(\left(\frac{1}{2}\right)^2-6\left(2-\frac{2}{3}\right)=-\frac{31}{4}=-7\frac{3}{4}\)

h(x)  = 6x + 1     we can write

y  =  6x + 1         the object  is to get x by itself and then "swap" x and y

Subtract 1 from both sides

y - 1  = 6x    divide both sides by 6

[ y - 1 ] / 6   =  x       "swap x and y

[x - 1 ] / 6  = y         for y, write  h^-1(x)

[ x - 1 ] / 6  =    h^-1 (x)         and this is the inverse

What are the vertical asymptotes of R(x) = 3x - 3/(x²) - 4?

As shown here, the vertcal asymptote occurs at x  = 0

https://www.desmos.com/calculator/uyregpotef

Area of 1st inner red  circle  = pi * ( diameter / 2) ^2  = pi * (2 / 2) ^2  =  pi (1)^2  = pi (1)

Area of 1st inner white ring  =    pi * ( 4/2 ) ^2  - (1)    = pi [ 4 - 1]  =  3pi    (2)

Area of 2nd red inner ring  =   pi * ( 6 / 2) ^2  -  [area of (1) + (2)]  =   9pi  - 3pi - pi  = 5pi      (3)

Area of 2nd inner white ring  =   pi * ( 8 / 2)^2  - [area of (1) + (2) + (3) ]   = 16pi  - 9pi  =  7 pi     (4)

Area of outer red ring   = pi * (10 /2 ) ^2  - area of  [(1) + (2) + (3) + (4)]  =  25 pi - 16pi  = 7pi

Sum of three red areas = pi  + 5pi +  7pi  = 13 pi  units^2  ≈  40.84 units^2

Area_circle  = pi * (diameter / 2) ^2    = pi * ( 18 /2 ) ^2  =   pi * ( 9)^2  =  81 pi   sq inches ≈  254.47 sq inches

(1/2)²−6(2−2/3)  =

1/4  - 6 ( 6/3 - 2/3)  =

1/4  - 6 ( 4/3)  =

1/4  -  24/3

1/4 - 8  =

1/4 - 32 / 4 =

-31 / 4 
 

 Simplify.
 

(1/2)²−6(2−2/3)
 

the answer to the question is 13 

what about part 1 boii

Percentage of area of circle=(area of circle/area of square)*100%

so,(pi(1)^2/2*2)*100% =78.54%

Let the area of smaller circle area be x and the larger circle be y

The larger circle have radius R ,than the smaller circle have radius R/2

so x/y =pi(R/2)^2 per pi(R)^2

equal 1/4

so,ratio x:y is 1:4

Relations and functions

If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find [ƒ(2) - g(1)] 2 <-- is to the power of two do i solve for fx and gx first and then use the power or do i multiply the brackets inside to the power first?

\([f(x)-g(x)]^2 \)

\(=[f(1)-g(2)]^2\)

\(=[(1^2+1)-(3\cdot2+1)]^2 \)

\(=[2-7]^2\)

\(=[-5]^2\)

\(=25\)

Do solve for fx and gx first and then use the power.

  !

Thanks Alan,

My answer is the same. :)

I'm just practicing :)

\(\frac{dp}{dt}=2\;cm/min\\ \frac{d\theta}{dt}=\frac{d\theta}{dp}\cdot \frac{dp}{d\theta}\\ \frac{d\theta}{dt}=\frac{d\theta}{dp}\cdot2\\ \frac{d\theta}{dt}=2\frac{d\theta}{dp}\\ \frac{d\theta}{dt}=2\times\frac{1}{6sec^2\theta}\\ \frac{d\theta}{dt}=\frac{cos^2\theta}{3}\\ \frac{d\theta}{dt}=\frac{36}{3q^2}\qquad \text{from diagram}\\ \frac{d\theta}{dt}=\frac{12}{(36+p^2)}\\~\\ When\;\; p=8\\ \frac{d\theta}{dt}=\frac{12}{(36+64)}\\ \frac{d\theta}{dt}=0.12\; radians/min \)

\(\dfrac{d}{dx}\dfrac{3}{4(2x+7)^9}\\ =\dfrac{3}{4}\dfrac{d}{dx}\dfrac{1}{(2x+7)^9}\\ =\dfrac{3}{4}\left(\dfrac{d}{dx}(2x+7)\right)\cdot\dfrac{-9}{(2x+7)^{10}}\\ =\dfrac{3}{4}\cdot 2\cdot -9\cdot \dfrac{1}{(2x+7)^{10}}\\ =\dfrac{-27}{2(2x+7)^{10}}\)

So there u go, with a denominator equals 2.

Speed = Distance/Time.

Therefore the average speed = x/p

\(xy + 3x^2y+1+2xy^2+yx^2+2+yx\\ =(xy + yx)+(3x^2y+yx^2)+2xy^2+(1+2)\\ =(xy + xy)+(3x^2y+x^2y)+2xy^2+3\\ =2xy + 4x^2y+2xy^2+3\)