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Assoli18 [71]
2 years ago
7

Write an equation in slope-intercept form for the line with slope2/5 and y-intercept -2.

Mathematics
2 answers:
andrezito [222]2 years ago
5 0

Answer

y=2/5x-2

Step-by-step explanation:

The slope-intercept equation for a straight line is y=mx+b,

where m is the slope and b is the y-intercept.

Here, m=2/5 and b=-2

⇒  y=−2/5x-2

Dimas [21]2 years ago
5 0

Answer:

the answer is

y =  \frac{2}{5} x - 2

Step-by-step explanation:

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Tcecarenko [31]

The cosine of an angle is the x-coordinate of the point where its terminal ray intersects the unit circle. So, we can draw a line at x=-1/2 and see where it intersects the unit circle. That will tell us possible values of θ/2.

We find that vertical line intersects the unit circle at points where the rays make an angle of ±120° with the positive x-axis. If you consider only positive angles, these angles are 120° = 2π/3 radians, or 240° = 4π/3 radians. Since these are values of θ/2, the corresponding values of θ are double these values.

a) The cosine values repeat every 2π, so the general form of the smallest angle will be

... θ = 2(2π/3 + 2kπ) = 4π/3 + 4kπ

b) Similarly, the values repeat for the larger angle every 2π, so the general form of that is

... θ = 2(4π/3 + 2kπ) = 8π/3 + 4kπ

c) Using these expressions with k=0, 1, 2, we get

... θ = {4π/3, 8π/3, 16π/3, 20π/3, 28π/3, 32π/3}

3 0
3 years ago
If x = 45, solve for y
Tanzania [10]

Answer:

d

Step-by-step explanation:

5 0
2 years ago
Use the zero product property to find the solutions to the equation x2 – 15x – 100 = 0.
IrinaK [193]

Answer:

x= 20      x =-5

Step-by-step explanation:

x^2 – 15x – 100 = 0.

What two numbers multiply to -100 and add to -15

-20 * 5 = -100

-20 +5 = -15

(x-20) (x+5) =0

Using the zero product property

x-20 =0     x+5 = 0

x= 20      x =-5

7 0
3 years ago
Read 2 more answers
Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

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Answer:

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Step-by-step explanation:

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