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2 years ago

What is the point-slope form of the equation for the line with a slope of -2 that passes through the point (4,-6)

Mathematics

2 answers:

LUCKY_DIMON [66]2 years ago

6 0

Answer:

$y + 6 = -2 (x-4)$

Step-by-step explanation:

Point-slope formula is represented by the equation $y-y_1 = m (x-x_1)$ . When knowing a point a line crosses through and its slope, you can write an equation in this form. Substitute the $m$ , $x_1$ , and $y_1$ for real values.

The $m$ represents the slope of the line. Since we know the slope is -2, substitute -2 for $m$ . The $x_1$ and $y_1$ represent the x and y values of the point the line passes through. Since the line passes through (4, -6), substitute 4 for $x_1$ and -6 for $y_1$ . This gives the following answer and equation:

$y-(-6) = -2 (x - (4))\\y + 6 = -2 (x-4)$

viva [34]2 years ago

3 0

Answer:

y=-2x+2

Step-by-step explanation:

y-y1=m(x-x1)

y-(-6)=-2(x-4)

y+6=-2(x-4)

y=-2x+8-6

y=-2x+2

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Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

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