Solve the system of linear equations by substitution, and show the steps please!

A sample of bacteria is being eradicated by an experimental procedure. The population is following a pattern of exponential deca

77julia77 [94]

Answer:

There will be 50 bacteria remaining after 28 minutes.

Step-by-step explanation:

The exponential decay equation is

$N=N_0e^{-rt}$

N= Number of bacteria after t minutes.

$N_0$ = Initial number of bacteria when t=0.

r= Rate of decay per minute

t= time is in minute.

The sample begins with 500 bacteria and after 11 minutes there are 200 bacteria.

N=200

$N_0$ = 500

t=11 minutes

r=?

$N=N_0e^{-rt}$

$\therefore 200=500e^{-11r}$

$\Rightarrow e^{-11r}=\frac{200}{500}$

Taking ln both sides

$\Rightarrow ln| e^{-11r}|=ln|\frac{2}{5}|$

$\Rightarrow {-11r}=ln|\frac{2}{5}|$

$\Rightarrow r}=\frac{ln|\frac{2}{5}|}{-11}$

To find the time when there will be 50 bacteria remaining, we plug N=50, $N_0$ = 500 and $r}=\frac{ln|\frac{2}{5}|}{-11}$ in exponential decay equation.

$50=500e^{-\frac{ln|\frac25|}{-11}.t}$

$\Rightarrow \frac{50}{500}=e^{\frac{ln|\frac25|}{11}.t}$

Taking ln both sides

$\Rightarrow ln|\frac{50}{500}|=ln|e^{\frac{ln|\frac25|}{11}.t}|$

$\Rightarrow ln|\frac{1}{10}|={\frac{ln|\frac25|}{11}.t}$

$\Rightarrow t= \frac{ln|\frac{1}{10}|}{\frac{ln|\frac25|}{11}.}$

$\Rightarrow t= \frac{11\times ln|\frac{1}{10}|}{{ln|\frac25|}}$

$\Rightarrow t\approx 28$ minutes

There will be 50 bacteria remaining after 28 minutes.

3 0

3 years ago

What is (7x-5)- (3x-2)

BaLLatris [955]

Answer:

4x-3

Step-by-step explanation:

Multiply the minus sign to lose the brackets

7x-5-3x+2

combine like terms

7x-3x-5+2

Simplify

4x-3 Answer

Hope this helps!

8 0

3 years ago

Can someone help me plz .. Use these lengths to find cos B, tan B, and sin B.

lara [203]

Answer:

cos B = $\frac{7}{25}$

tan B = $\frac{24}{7}$

sin B = $\frac{24}{25}$

Step-by-step explanation:

In the right triangle, there are three sides and 2 acute angles

Hypotenuse ⇒ the opposite side of the right angle

Leg1 and Leg 2 ⇒ the sides of the right angle

The trigonometry functions of one of the acute angles Ф are

sin Ф = opposite leg/hypotenuse

cos Ф = adjacent leg/hypotenuse

tan Ф = opposite leg/adjacent leg

In Δ ACB

∵ ∠C is the right angle

∴ AB is the hypotenuse

∵ AC is the opposite side of ∠B ⇒ leg1

∵ CB is the adjacent side of ∠B ⇒ leg2

→ By using the ratios above

∴ cos B = $\frac{CB}{AB}$ , tan B = $\frac{AC}{CB}$ , sin B = $\frac{AC}{AB}$

∵ CB = 7, AB = 25, AC = 24

∴ cos B = $\frac{7}{25}$

∴ tan B = $\frac{24}{7}$

∴ sin B = $\frac{24}{25}$

6 0

3 years ago

3 3/5 + 6 3/10 please

Dafna11 [192]

In order to solve first convert both values into improper fractions:
18/5+63/10
second, convert both values to fractions with a base of ten:
36/10+63/10
finally, add the numerators:
99/10 is the final answer

3 0

3 years ago

How do i graph the lines when the equations aren’t in slope intercept form????

jekas [21]

Slope-Intercept Form: y=mx+b

Standard Form: ax+by=c

Point- Slope: (y-y1)= m(x-x1)

There are multiple answers to your question-

If you are only missing b(the y-intercept) but are given a set of points, plug the points into x and y and solve for b.
If you are only missing the slope(m) but are given a set of points, plug the points into x and y and solve for m.
If you are given the standard form/point-slope form, change the equation to slope intercept form.
If you are given an complete form(there is an x and y; no missing variables), but are not sure what it is, plug in some numbers in x to find y, then graph.

7 0

3 years ago