hectictar

$f(x)\ =\ 6\sqrt{3x}\\~\\ g(x)\ =\ (2x)^4\ =\ 2^4x^4\ =\ 16x^4\\~\\ h(x)\ =\ \big(2^{x+2}\big)^2\ =\ \big(2^{2}\big)^{x+2}\ =\ 4^{x+2}\ =\ 4^2\cdot4^x\ =\ 16\cdot4^x\\~\\ j(x)\ =\ 3\sqrt{12x}\ =\ 3\sqrt{4\cdot3x}\ =\ 3\sqrt{4}\sqrt{3x}\ =\ 6\sqrt{3 x}\\~\\ m(x)\ =\ 16x^4$

So we can see that   $f(x)\ =\ j(x)$   and   $g(x)\ =\ m(x)$

( And  $h(x)$  is not equal to any of the other functions. )

$f(x)\ =\ 2^{x+3}\ =\ 2^x\cdot2^3\ =\ 2^3\cdot2^x\ =\ 8\cdot2^x\ =\ 8\cdot g(x)\\~\\ f(x)\ =\ 8\cdot g(x)$_

25x² + 20x - 10  =  0

                                       Rewrite the equation as...

(5x)² + 4(5x) - 10  =  0

                                       To make it clearer, let  u  =  5x   and then substitute  u  in for  5x

u² + 4u - 10  =  0

                                       Add  10  to both sides of the equation.

u² + 4u  =  10

                                       Add  4  to both sides of the equation to complete the square on the left side.

u² + 4u + 4  =  14

                                       Factor the perfect square trinomial on the left side.

(u + 2)²  =  14

                                       And since  u  =  5x  we can substitute  5x  in for  u

(5x + 2)²  =  14

Now the equation is in the form  (ax + b)²  =  c   where  a,  b,  and  c  are integers and  a > 0 .

a + b + c  =  5 + 2 + 14  =  21

If the quadratic has exactly one real root then the discriminant must be zero.

(4m)² - 4(1)(m)  =  0

16m² - 4m  =  0

m(16m - 4)  =  0

16m - 4  =  0          or          m  =  0

16m  =  4

m  =  1/4

The positive value of  m  is  1/4

Let's say it starts wih  x  fish.

The smallest positive integer value of  x  that makes  $\frac{8}{27}x-\frac{38}{27}$  a positive integer is  25 .

Hahaha!!

The endpoints of  s₂  are  3  to the right and  2  down from the endpoints of  s₁

s₁  has endpoints at  (1, 2)  and  (7, 10)

s₂  has endpoints at  (1 + 3, 2 - 2)  and  (7 + 3, 10 - 2)

s₂  has endpoints at  (4, 0)  and  (10, 8)

point M  =  midpoint of segment PR  =  $\Big( \frac{1+7}{2},\frac{3+15}{2}\Big)\ =\ \Big( \frac{8}{2},\frac{18}{2}\Big)$  =  (4, 9)

To reflect segment PR over the x-axis, we make the y-coordinate of each of its points negative. So...

image of point M  =  (4, -9)

sum of the coordinates of the image of point M  =  4 + -9  =  -5

circumference of base of cone  =  length of major arc AB

length of major arc AB  =  $\frac{270}{360}\cdot$ circumference of circle

length of major arc AB  =  $\frac{270}{360}\cdot2\pi\cdot4"$

length of major arc AB  =  $\frac{3}{4}\cdot2\pi\cdot4"$

length of major arc AB  =  $6\pi"$_

tan( angle )  =  opposite / adjacent

tan( 30° )  =  9√3 / PQ

PQ tan( 30° )  =  9√3

PQ  =  9√3 / tan( 30° )

PQ  =  27

....Or.....you can notice that this is a 30-60-90 triangle. And remember in every 30-60-90 triangle,

side opposite the 60° angle  =  √3 * ( side opposite the 30° angle )

PQ  =  √3 * 9√3

PQ  =  3 * 9

PQ  =  27