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2 years ago

Passes through (2, 3) at a gradient of 2

Mathematics

1 answer:

Tcecarenko [31]2 years ago

6 0

Answer:

y = 2x-1

Step-by-step explanation:

I'm guessing gradient of 2 means slope of 2

if line passes through (2,3) and has a slope of 2, we can use the slope intercept equation

slope-intercept (y=mx+b)

m is the slope and b is the y-intercept (which we are trying to find)

so far, we have a point and a slope so we can fill in x,y, and m

(3) = (2)(2) + b

3 = 4 + b

3-4 = 4-4 + b (to undo addition, you use subtraction so I minus 4 from each side)

-1 = b

we found b so we just take the (2,3) out of the equation substitute the x,y back into the equation

y = 2x - 1

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1. Contaminated water is subjected to a cleaning process. The concentration of the

Leokris [45]

Answer:

The concentration of pollutants in the water can be defined as the quotient between the mass of pollutants and the total mass of water.

Assuming that 1L of water weighs 1000 grams, we know that:

Mass of pollutants = 5mg = 0.005g.

Initial concentration = 0.005g/1000g = 0.000005

If we multiply this by 100%, we get the percentage:

0.000005*100% = 0.0005%

Now, the mass of pollutants decreases by 10% each hour.

So if initially, we have 5mg

After one hour, we will have: 5mg - 0.1*5mg = 5mg*(0.9).

After another hour, we will have: 5mg*(0.9) - 0.1*5mg*0.9 = 5mg*(0.9)^2.

And so on, then after n hours, the mass of pollutants will be:

M(n) = 5mg*(0.9)^n

Or we can write this in grams as:

M(n) = 0.005g*(0.9)^n

Then the concentration as a function of time in hours will be:

C(n) = M(n)/1000g = (0.005g/1000g)*(0.9)^n

Notice that the thing between the parentheses is the initial concentration, then if we write this in percentage form:

c(n) = 100%*(0.005g/1000g)*(0.9)^n = 0.0005%*(0.9)

The function that represents the concentration of polution in the water as a function of hours is:

c(n) = 0.0005%*(0.9)

6 0

3 years ago

Find the distance between P1(4,16degrees) and P2(-2,177degrees) on the polar plane.

bazaltina [42]

Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.

You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.

Let's convert (4, 16°) to Cartesian coordinates.

Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375

sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327

So (4, 16°) ⇒ (3.84504678375, 1.10254942327).

Let's convert (-2, 177°) to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)

In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951

sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248

So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}$
$d\approx\sqrt{-1.84778771^2+-1.20722134^2}$
$d\approx\sqrt{3.41431942+1.45738336}$
$d\approx\sqrt{4.87170278}$
$\boxed{d\approx2.20719342}$

7 0

3 years ago

SOLVED

Nastasia [14]

5)
a. The equation that describes the forces which act in the x-direction:
 Fx = 200 * cos 30 

b. The equation which describes the forces which act in the y-direction: 
 Fy = 200 * sin 30 

c. The x and y components of the force of tension: 
 Tx = Fx = 200 * cos 30 
 Ty = Fy = 200 * sin 30 

d.Since desk does not budge, frictional force = Fx
= 200 * cos 30 

 Normal force = 50 * g - Fy
= 50 g - 200 * sin 30
____________________________________________________________
6) Let F_net = 0
a. The equation that describes the forces which act in the x-direction:
(200N)cos(30) - F_s = 0

b. The equation that describes the forces which act in the y-direction:
F_N - (200N)sin(30) - mg = 0

c. The values of friction and normal forces will be:
Friction force= (200N)cos(30),

The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N

Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N

4 0

3 years ago

Can u help??????????????????????????????????????//

maw [93]

Answer:

Okay!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The answer is 14

Step-by-step explanation:

Folow order of operations and when doing multipactiona or subtraction or division and addition, make sure to go from left to right

5 0

3 years ago

Read 2 more answers