Least common multiple of 6,5,3

Solve to three significant digits 123= 500e ^ -0.12x

Anon25 [30]

$500{ e }^{ -0.12x }=123\\ \\ { e }^{ -\frac { 12 }{ 100 } x }=\frac { 123 }{ 500 } \\ \\ { e }^{ -\frac { 3 }{ 25 } x }=\frac { 123 }{ 500 }$

$\\ \\ { \left( { e }^{ x } \right) }^{ -\frac { 3 }{ 25 } }=\frac { 123 }{ 500 } \\ \\ { e }^{ x }={ \left( \frac { 123 }{ 500 } \right) }^{ \frac { 1 }{ -\frac { 3 }{ 25 } } }$

$\\ \\ { e }^{ x }={ \left( \frac { 123 }{ 500 } \right) }^{ -\frac { 25 }{ 3 } }\\ \\ \ln { \left( { \left( \frac { 123 }{ 500 } \right) }^{ -\frac { 25 }{ 3 } } \right) } =x$

$\\ \\ x=-\frac { 25 }{ 3 } \ln { \left( \frac { 123 }{ 500 } \right) }$

$\therefore \quad x\approx 11.7$

8 0

3 years ago

Graphing the function. h(x) = - 4 (show steps)

Archy [21]

Think about rise over run. Make the -4 a fraction into -4/1. Rise is the first, so you go down 4 pts down. Then run is 1, so you go 1 right. In this case, there is no beginning, so you begin at 0. The coordinates will be (0, 0) and (1, -4). You got the graph.

3 0

3 years ago

A company has an opportunity to bid on three contracts. Determine which would be the best investment given the information in th

MariettaO [177]

CONTRACT PROFIT, (P)* OF PROFIT...(P)*. TO BREAK EVEN...LOSS, (P)* OF LOSS

Southeast $45,000, 50% 30% $6,000, 20%

Southwest $60,000, 35% 40% $10,000, 25%

California $112,000, 20% 40% $40,000, 40%

I think that the best contract would be with SOUTHEAST. It probability of having a profit is higher than the other two at 50% while its probability of incurring a loss is lesser than the other two at 20%.

Despite the bigger value of its contract profit of the other two, these value will still be greatly affected by its corresponding probabilities.

6 0

3 years ago

Read 2 more answers

Expand the following x(x-6), x(4x-1), 2x(5x+4), 3x(5x-y)

zlopas [31]

Answer:

x^2-6x, 4x^2-1x,10x^2+8x,15x^2-3xy

Just use the distributive property like you have before and you will get these answers, Hope this helped!

4 0

3 years ago

Read 2 more answers

Ex 2.8<br> 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1

harkovskaia [24]

$y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\
y''(0)=20\cdot0^3=0$

The value of the second derivative for $x=0$ is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of $5x^4$ is always positive for $x\in\mathbb{R}\setminus \{0\}$ . That means at $x=0$ there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval $[-2,1]$ .
The function $y$ is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

$y_{max}=y(1)=1^5-3=-2$

4 0

3 years ago