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3 years ago

HELP ASAP GEOMETRY . . . . . .

Mathematics

1 answer:

lukranit [14]3 years ago

6 0

Answer:

1. y = -(2/5)x - (1/5)

2. y = -(9/5)x - 4

Step-by-step explanation:

For 1:

Step 1: rewrite the equation of the given line in to slop-intercept form by solving for y

2x + 5y = 15

5y = -2x + 15 (subtract 2x from both sides)

y = -(2/5)x + 3 (divide both side by 5)

Step 2: Our line is parallel to this line, so it has the same slope, but a different y-intercept, so set up the equation...

y = -(2/5)x + b

We are given a point (x, y) of (2, -1), so plug that in and solve for b.

-1 = -(2/5)(2) + b

-1 = -4/5 + b (simplify)

4/5 -1 = b (add 4/5 to both sides to isolate b)

4/5 - 5/5 = b

-1/5 = b

So the equation of our line is y = -(2/5)x - (1/5)

For 2:

Step 1: Perpendicular lines have slopes that are opposite reciprocals of each other. That means you take the slope, flip the fraction, and change the sign.

Here our given line is y = (5/9)x - 4 so the slope 5/9

The opposite reciprocal of 5/9 is -9/5

We set up the equation

y = -(9/5)x + b

we are given a point (x, y) of (-5, 5), so plug that in and solve for b

5 = -(9/5)(-5) + b

5 = 9 + b

-4 = b

So our equations is y = -(9/5)x - 4

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Determine the value of w and x in the figure below. (Thanks for the help I really appreciate it!)

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4 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

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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<h2>✏Part B :- </h2><h2 /><h2>The Earth could look like within 100 years if we succeed in curbing climate change.</h2>

<h2>Hope it helps.❤</h2><h2 />

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