hectictar

\(\quad\sqrt{48x^4y^3} \\ =\\ \quad\sqrt{2\cdot2\cdot2\cdot2\cdot3\cdot x^4\cdot y^3} \\ =\\ \quad\sqrt{2^4\cdot3\cdot x^4\cdot y^3} \\ =\\ \quad\sqrt{2^4\cdot3\cdot x^4\cdot y^2\cdot y} \\ =\\ \quad\sqrt{2^4\cdot x^4\cdot y^2\cdot 3\cdot y} \\ =\)

\(\quad\sqrt{2^4}\cdot \sqrt{x^4}\cdot \sqrt{y^2}\cdot \sqrt{3\cdot y}\\ =\\ \quad2^2\cdot x^2\cdot y\cdot\sqrt{3\cdot y}\\ =\\ \quad4x^2y\sqrt{3 y} \)          Since  x  and  y  are positive, we can say...

population after  0  hours  =  50

population after  18  hours  =  2 * 50

population after  2*18 hours  =  2 * 2 * 50

population after  3*18 hours  =  2 * 2 * 2 * 50  =  2^3 * 50

population after  x*18 hours  =  2^x * 50

x*18  =  h

x  =  h/18

population after  h  hours  =  2^(h/18) * 50

population after  60  hours  =  2^(60/18) * 50   ≈  504

The question is...

What are the coordinates of the point on the graph of  f(x)  where the slope of f(x)  =  -3  ?

the slope of f(x) at any  x  value  =  f'(x)  =  -2x + 3

What  x  value makes the slope of f(x) be  -3  ?

What  x  value makes  f'(x)  be  -3 ?

f'(x)  =  -3

-2x + 3  =  -3

-2x  =  -6

x  =  3

When  x = 3 ,  the slope of f(x)  =  -3

When  x = 3 ,  f(x)  =   f(3)  =  2

So at the point  (3, 2) , the slope of  f(x)  is  -3 .

\(h=\sqrt{\frac{g^2}{d}-5t}-1\)

                                          Add  1  to both sides of the equation.

\(h+1=\sqrt{\frac{g^2}{d}-5t}\)

                                          Square both sides of the equation.

\((h+1)^2=\frac{g^2}{d}-5t\)

                                          Add  5t  to both sides.

\((h+1)^2+5t=\frac{g^2}{d}\)

                                                Multiply both sides by  d .

\(d[(h+1)^2+5t]=g^2\)

                                                Take the ± square root of both sides.

\(±\sqrt{d[(h+1)^2+5t]}=g \\~\\ g=±\sqrt{d[(h+1)^2+5t]}\)

P(x)  is the population of rabits at any point in any year. To get the population after  3  years, calculate  P(3). To get the population halfway through the third year, calculate  P(3.5) .

P'(x)  is the instantaneous rate of change in population per change in year. For example...

P'(11) = -4  means that after  11  years, the population was increasing by  -4  per  1 year....in other words....the population was decreasing by  4  per year.

x  =  number of round tables     and     y  =  number of rectangular tables

Each round table has 6 chairs. Each rectangular table has  10  chairs.

The total number of chairs must be greater than or equal to  185  .

6x + 10y  ≥  185

There must be no more than  15  rectangular tables.

So the number of rectangular tables must be less than or equal to  15 .

y  ≤  15

There must be no fewer than  6  round tables.

So the number of round tables must be greater than or equal to  6 .

x  ≥  6

The number of rectangular tables must be greater than or equal to one-third the number of round tables.

y  ≥  \(\frac13\)x

3y  ≥  x

x  ≤  3y

First box:     x⁴ - 2x²y² + y⁴

Second box:     2x⁴ - 2x²y² + 2y⁴

Third box:     2(x⁴ - x²y² + y⁴)

g(x)   =   2^{x - 3} + 2   =   2^{x - 6 + 3} + 3 - 1   =   (2^{(x + 3) - 6} + 3) - 1   =   f(x + 3) - 1

So

the graph of g(x)  is the graph of  f(x)  shifted  3  units left and  1 unit down.

2.   (tan²x) / (1 + sec x)   =   sec x + 1      Let's test whether this is true when  x = 0 .

(tan²x) / (1 + 1/cosx)   =   1/cosx + 1

(tan²0) / (1 + 1/cos0)   =   1/cos0 + 1

(0²) / (1 + 1/1)   =   1/1 + 1

0  =  2       This is false, so this equation is not true for all values of  x  and so it is not an indentity.

If you want to find the solutions to the equation, we can try to find them.

(tan²x) / (1 + sec x)   =   sec x + 1       Multiply both sides by  (1 + sec x)

tan²x   =   (sec x + 1)(1 + sec x)

tan²x  =  sec x + sec²x + 1 + sec x

tan²x  =  sec²x + 2sec x + 1                Rewrite  tan  and  sec  in terms of  sin  and  cos.

sin²x / cos²x  =  1/cos²x + 2/cos x  +  1        Multiply through by  cos²x

sin²x  =  1 + 2cos x + cos²x                 Subtract  sin²x  from both sides.

0  =  1 + 2cos x + cos²x - sin²x           Substitute  1 - cos²x  in for  sin²x

0  =  1 + 2cos x + cos²x - (1 - cos²x)

0  =  1 + 2cos x + cos²x - 1 + cos²x

0  =  2cos x + 2cos²x         Divide through by  2 .

0  =  cos x + cos²x             Factor  cos x  out of both terms.

0  =  cos x(1 + cos x)         Set each factor equal to zero.

cos x  =  0      or      1 + cos x  =  0

However, neither of these can be true.

If cos x  =  0  then  sec x  is undefined and the original equation is undefined.

If  cos x  =  -1  then  1 + sec x  =  1 - 1  =  0  and the original equation is undefined.

So this equation has no solutions.

1.   sin²(x) - 2  =  -sin(x)

This can't be an identity because it is not always true. For instance...

sin²(0) - 2  =  -sin(0)

0² - 2  =  -0

-2  =  0        This is false, so this equation is not an identity.

However, this equation does have solutions. We can find its solutions.

sin²(x) - 2  =  -sin(x)

                                             Add  sin(x)  to both sides of the equation.

sin²(x) + sin(x) - 2  =  0

                                             Factor the left side like this:  u² + u - 2  =  (u + 2)(u - 1)

(sin(x) + 2)(sin(x) - 1)  =  0

                                                             Set each factor equal to zero.

sin(x) + 2  =  0     or     sin(x) - 1  =  0

sin(x)  =  -2          or     sin(x)  =  1

sin  returns values between  -1  and  1  inclusively, so we can rule out the first option. That leaves...

sin(x)  =  1

x  =  arcsin(1)  =  90°

There will also be a solution every time  360°  is added to  90° , so all the real solutions are

x  =  90° + 360°n      where  n  is an integer

Does this answer your question? I might have misunderstood your question.