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3 years ago

Linear equations. write the standard form of an equation of the

Mathematics

1 answer:

erastova [34]3 years ago

7 0

Answer:

The standard form of an equation of the line that passes through the given point having slope is $y-y_{1} =m(x-x_{1}$

Step-by-step explanation:

The standard form of an equation of the line that passes through the given point ( $(x_{1} ,y_{1} )$ and having slope m is

$y-y_{1} =m(x-x_{1}$

another form of an equation of the line that passes through the given point ( $(x_{0} ,y_{0} )$ and having slope m is

$y-0 =m(x-0)$

y=m x

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Find the product 9x2x7

Tema [17]

Answer:

126

Step-by-step explanation:

Since 9 times 2 is eighteen, and 18 times 7 is 126, the answer is 126.

8 0

3 years ago

Read 2 more answers

Please answer correctly !!!!! Will mark brainliest !!!!!!!!!!

Sever21 [200]

Answer:

Exactly one solution

Step-by-step explanation:

4x -2y = 8

2x +y = 2

Multiply the second equation by 2

4x +2y = 4

Add the two equations together

4x -2y = 8

4x +2y = 4

-------------------

8x = 12

x = 12/8

We know there will be one solution

4 0

3 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago

1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+10 times 2

Maksim231197 [3]

Answer:

200

Step-by-step explanation:

Step 1: Add

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

1 + 2 = 3

3 + 3 = 6

6 + 4 = 10

10 + 5 = 15

15 + 6 = 21

21 + 7 = 28

28 + 8 = 36

36 + 9 = 45

45 + 1 = 46

46 + 2 = 48

48 + 3 = 51

51 + 4 = 55

55 + 5 = 60

60 + 6 = 66

66 + 7 = 73

73 + 8 = 81

81 + 9 = 90

90 + 10 = 100

100

Step 2: Multiply

100 * 2

200

Answer: 200

6 0

3 years ago

60 - 3d + 20 - 6 = 4d

iragen [17]

60-3d + 20-6 =4d
74-3d =4d
-3d =4d -74
-7d = -74
D=74/7

4 0

3 years ago

Read 2 more answers