16 - (-9 - -8)= ????

A line passes through (-7,-5) and (-5,4). Write an equation for the line in point-slope form. Rewrite the equation in standard f

rewona [7]

The equation:
y - y 1 = ( y2 - y1 ) / ( x2- x1) * ( x - x1 )
y - ( - 5 ) = ( 4 + 5 ) / ( - 5 + 7 ) * ( x - ( - 7 ) )
y + 5 = 9/2 ( x + 7 )
y + 5 = 9/2 x + 63 /2 / * 2
2 y + 10 = 9 x + 63
- 9 x + 2 x = 53
Answer:
C ) y + 5 = 9/2 ( x + 7 ) ; - 9 x + 2 y = 53

4 0

3 years ago

Read 2 more answers

The sum of the numbers from 1 to n can be performed recursively as follows: The sum from 1 to 1 is 1. The sum from 1 to n is n m

pogonyaev

Answer:

S(1) = 1

S(n) = S(n-1) + n

Where S(n) is the sum from 1 till n

And,

S(n-1) is the sum from 1 till n-1

4 0

3 years ago

PLEASE HELP I NEED TO KNOW THE (RIGHT) ANSWER ASAP!

shtirl [24]

Triangles are 180 degrees total so 80+40=120, subtract that from 180 and you’ll get B=60

4 0

3 years ago

Read 2 more answers

Find the sum of the first 50 terms in the sequence where a1=2.1 and a50=1.2

Elan Coil [88]

Answer:

82.5 (assuming it is an arithmetic sequence)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

3 years ago