How do i solve x² + x - 2

Mathematics

1 answer:

jeka943 years ago

8 0

Square x first then add x then subtract by 2

You might be interested in

Cubed means an exponent of 2. true or false

stepan [7]

Answer:

This is false

Step-by-step explanation:

Cubed is an exponent of 3, a cube is a 3D shape.

Exponent of 2 is squared, s square is a 2D shape.

7 0

2 years ago

Pls answer 18 20 and 22 show your work

Svetlanka [38]

Answer:

18- 300

20- 1,500

22- 300

Step-by-step explanation:

You can easily find the answer like this:

18. Multiply 60 and 100 together. The hundred is from 100%. After that, divide it by 20 because of the percent. After doing this, you should get 300.

20. 360*100=36,000 36,000/24= 1,500

22. 9*100=900 900/3=300

I hope this is what you were looking for. If you need more out of this, ask me to answer it again. :)

5 0

3 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$