hectictar

The side across from the 60° angle is √3 times the side across from the 30° angle,

and the side across from the 30° angle is half the hypotenuse.  So...

a   =   8/2 * √3   =   4√3

b   =   a/2 * √3   =   4√3 / 2  *  √3   =   2√3 * √3   =   2 * 3   =   6

c   =   b/2 * √3   =   6/2 * √3   =   3√3

d   =   c/2   =   3√3 / 2

(where a, b, c, and d are all in cm)

The graph of  y = x² + bx + c  is a parabola that opens upward and intersects the x-axis at  -4  and  8 .

An equation for a parabola that opens upward and intersects the x-axis at  -4  and  8  is  y = (x + 4)(x - 8)

Let's take the second equation and expand the right side by FOILing it:

y  =  (x + 4)(x - 8)

y  =  x² - 4x - 32

Now we can see that  x² - 4x - 32  matches  x² + bx + c  and so we can say...

b = -4

c = -32

b + c = -36

How can the first book and the last book both be math books if all math books must be at the beginning?

a)

\(\text{total pizza ate}\ =\ \frac13\ \text{of a pizza}\ +\ \frac14\ \text{of a pizza} \\~\\ \text{total pizza ate}\ =\ (\frac13+\frac14)\ \text{of a pizza} \\~\\ \text{total pizza ate}\ =\ (\frac{1\times4}{3\times4}+\frac{1\times3}{4\times3})\ \text{of a pizza} \\~\\ \text{total pizza ate}\ =\ (\frac{4}{12}+\frac{3}{12})\ \text{of a pizza} \\~\\ \text{total pizza ate}\ =\ \frac{7}{12}\ \text{of a pizza} \)

b)

Technicallyyyy there could be any number of slices in the pizza. But 12 would be a good number of slices. That way Khalid could eat 4 full slices and his sister could eat 3 full slices.

\((a+b)\sqrt{cd}\)       Replace  a  with  2 ,  b  with  4 ,  c  with  3 ,  and  d  with  12

\(=\ (2+4)\sqrt{(3)(12)}\)       Now we can simplify this expression

\(=\ (6)\sqrt{36}\)

\(=\ (6)(6)\)

\(=\ 36\).

B)

f(x)  =  √x

Then  f  is not differentiable at  x = 0  because  f'(x)  =  \(\frac12x^{-\frac12}=\frac{1}{2\sqrt{x}}\)   which is undefined when  x = 0

And

f²(x)  =  ( √x )²   =   x

Which is differentiable at  x = 0  because  f²'(x)  =  1  for all values of  x .

A)

Let   f₁(x)  =  x² + 1   and let   f₂(x)  =  kx

In order for  f(x)  to be differentiable at  x = 1, two things must be true at the same time:

First, the graphs must "touch" at  x = 1.  That is,   f₁(1)  =  f₂(1)

Second, the slopes of both graphs at the point where they touch must be identical.  That is,   f₁'(1)  =  f₂'(1)

We can start with either condition and find which value of  k  makes it true, and then check to make sure that same value of  k  makes the other condition true. Let's start with the first condition:

f₁(1)  =  f₂(1)

                            And     f₁(1)  =  1² + 1     and     f₂(1)  =  k(1)

1² + 1  =  k(1)

                            Simplify both sides of the equation.

2  =  k

So we know that  k = 2  satisfies the first condition. Now let's check to make sure it also satisfies the second condition:

f₁'(1)  =  f₂'(1)

                            f₁'(x)  =  2x     so     f₁'(1)  =  2(1)  =  2     and     f₂'(x)  =  k  =  2

2  =  2

The second condition is also satisfied when  k = 2.

So     k = 2

I think it is the opposite guest... IDK the best way to explain it, but look at this graph:

What percent of 24 is 80?

x%  of  24   =   80

                                       Replace  %  with  /100   and   replace  "of"  with  *

x/100  *  24   =   80

                                       Divide both sides of the equation by  24

x/100   =   80/24

                                       Multiply both sides of the equation by 100

x   =   80/24 * 100

                                       Calculate  80/24 * 100

x   ≈   333.33

So.....

333.33%  of  24  is  80.

6j² - 6j - 12

                          Factor  6  out of all the terms

=   6(j² - j - 2)    (Notice here that if we distributed the 6, we would get back the previous expression.)

Now we want to force there to be a perfect square trinomial within the parenthesees. To do that, let's add and subtract half the coefficient of the middle term squared. That is, let's add and subtract  (1/2 * -1)²  which is  (-1/2)²  which is  1/4 . This will create a perfect square trinomial without changing the value of the expression.

=   6(j² - j + \(\frac14\) - \(\frac14\) - 2)

                                       Now there is a perfect square trinomial which is highlighted below:

=   6(j² - j + \(\frac14\) - \(\frac14\) - 2)

                                       Then  j² - j + 1/4  factors as  (j - 1/2)²

=   6( (j - \(\frac12\))²  - \(\frac14\) - 2)

                                       And  2 = 8/4

=   6( (j - \(\frac12\))²  - \(\frac14\) - \(\frac84\))

                                       And -1/4 - 8/4  =  -9/4

=   6( (j - \(\frac12\))²  - \(\frac94\) )

                                       Distribute  6  to the terms in parenthesees

=   6( j - \(\frac12\) )² -  6( \(\frac94\) )

=   6( j - \(\frac12\) )² -  \(\frac{27}{2}\)

Now it is in the form  c(j + p)² + q,  where   c = 6,  p =  -\(\frac12\),  and  q = -\(\frac{27}{2}\)

So  q / p   =   ( -\(\frac12\) ) / ( -\(\frac{27}{2}\) )   =   ( -\(\frac12\) ) * ( -\(\frac{2}{27}\) )   =   \(\frac{1}{27}\)