All Answers 357307 Answers

The sum of all values of x such that f(x) = 0 is -3 + (-2) = -5.

We need x^2 - 3x - 4 to be nonzero, so the domain is (-inf,-1) U (-1,4) U (4,inf).

I'm completly stumped. 

Perhaps one of the smarter people on the forum can help answer?

If you manage to figure it out, please tell me.

Sorry

=^._.^=

Hello! Thank you, I understand that part! However, one question I have is that for some sqrtb^2-4ac that are integers still don't give out integer solutions because they are fractions. How would I know which ones work or not?

Hi Guest!

I have yet to learn about ellipses, so I'm sorry I can't help you on problems 2a or 2b.

However, I'm going to try question 1. 

The quadratic equation is (-b+-sqrt(b^2-4ac))/2a, where the +- is the plus/minus weird symbol. 

For the roots to be integers, sqrt(b^2-4ac) must be an integer. 

b = -8

c = 3

sqrt(64-12a) = integer 

2sqrt(16-3a) = integer

Can you try solving for them?

=^._.^=

Hi Guest!

Try to perhaps screenshot problems in the future so it's easier to read. 

If LM is 15, then the circufremce of the circle is 30pi. 

60 degrees is 1/6 of the cricle, so the answer is 30pi/6 = 5pi. 

5pi is about 15.71

I hope this helped. :)))

=^._.^=

There are 3 values of x such that f(f(x)) = 5.

A + B + C = 18.

The smallest possible integer value of x is 3.

Nice solution!!!

Connect the centers of the circles. You will get a line of length 2r.

You can also draw a line starting at the center higher circle going downward and a line starting at the center of the lower circle to form an isosceles right triangle with hypotenuse 2r.

The side length of a leg of the isosceles right triangle is $r\sqrt{2}$.
Therefore, the side length (s) = $2r + r\sqrt2$.

Let's find the lowest $2+4x^2$ can be and the highest.
The lowest is when x = 0, and the value of the fraction is 1. When the value of the denominator is the lowest, the value of the fracture is highest (I hope that makes sense) so the highest value of the fraction is 1. (So b = 1)

The highest is .... well, there is no highest. However, no matter how high the denominator is, 2/(big number) the fraction will never be 0. So a = 0.

We have a = 0, and b = 1, so the sum is 1.

7 ( x² - 9 )

7 (x-3)(x+3)

The answer is 18. I would explain how I got it, but I just used trial and error.

s in this case = 2x from 0 to 3

  should be      s in this case = 2 y from 0 to 3

a)    put '3' in for 't'  in the equation given and calculate the height at 3 sec

b.)   max height will occur at t = -b/2a    where b = 64   and a = -16

          this is the TIME when the apple is at max height

              to calculate THIS HEIGHT , put the t you just found into the equation and calculate max height

c) When the apple becomes applesauce, the height = 0

         so   solve   0 = -16t² + 64 t + 80         use Quadratic Formula or factoring       throw out any negative 't' value     

5 cube sides      5 *   7*7

4 triangles   area = 1/2 b h   =    4   *  1/2 7*8

Add the red equations to get total surface  area in cm²

The question is as follows:

The base of a solid is the region between the parabolas x=y^2 and 2y^2 = 3-x . Find the volume of the solid if the cross-sections perpendicular to the x-axis are equilateral triangles.

Hello Guest!

The parabolas intersect in P(1,1) and P(1,-1). They share the domain of the integral.

The entire domain is $0\le x\le3$.

$A_{\Delta}=\frac{a^2\sqrt{3}}{4}$       

$\large\frac{k^2\sqrt{3}}{4}\\ \Rightarrow \ k_0^1=2\cdot f(x)\\ \Rightarrow \ k_1^3=2\cdot g(x)$

${\color{black}x=y^2\\ \Rightarrow} \ f(x)=\sqrt{x}\\ {\color{black}2y^2=3-x\\ \Rightarrow} \ g(x)=\sqrt{\frac{3-x}{2}}$

$V=\int_{0}^{1} \! (2\cdot f(x))^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot g(x))^2\cdot \frac{\sqrt{3}}{4} \, dx$

$V=\int_{0}^{1} \! (2\cdot \sqrt{x})^2\cdot \frac{\sqrt{3}}{4} \, dx +\int_{1}^{3} \! (2\cdot \sqrt{\frac{3-x}{2}})^2\cdot \frac{\sqrt{3}}{4} \, dx$

$V=\sqrt{3}(\int_{0}^{1} \! x \, dx +\frac{1}{2}\int_{1}^{3} \! (3-x) \, dx )$

$V=\sqrt{3}\cdot (|\frac{x^2}{2}|_0^1+|3x-\frac{x^2}{2}|_1^3)\\ V=\sqrt{3}\cdot (\frac{1}{2}+9-\frac{9}{2}-(3-\frac{1}{2})$

$V=\frac{5\cdot \sqrt{3}}{2}=4.33$

  !

speed= wavelength * freq

340  = wl * 540

340/540 = wavelength in meters                       (I think)

a₁ = -28       d = -13

a₇₃  =  -28  +  72 (-13) = ...........

I'll TRY.....First here is a graph of the area of the base:

  area of equilateral triangle =  sqrt(3)/4 s^2

s in this case = 2x from 0 to 3

 from 0 to 1   this is    2y^2            from  1 to 3 this is     6-4y^2      now square s

                                  4 y⁴                                                     16y^4 -48y^2 + 36

sqrt( 3)/4  ( 4y^4 )                             +          sqrt (3)/4 ( 16 y^4 - 48y^2 + 36)   

      this part is y 0 to 1                                          this part is y is 1 to 0

Hmmmm...... I am a little stuck...a bit rusty......If I figure this out (this may not be the EXACT way to do it) ...I'll edit this post

    maybe asinus will answer correctly....

d = 5

a₁ = 16

a₉₀  = a₁    = 89 d

        = 16 + 89 (5) = .........

2/3 * batch = 4 cups

batch = 4 * 3/2   cups = ............   cups

2/5 * c0st = 240

cost =     240 * 5/2   = $ ..........