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Triss [41]
2 years ago
10

What is the equation of the line that passes through the points (4,-1) and (3,-8)

Mathematics
1 answer:
Minchanka [31]2 years ago
4 0

Answer:

y=7x-29

Step-by-step explanation:

used a graphing calculator

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Solve.
Serggg [28]

Answer:

The solution is (4, 0)

Step-by-step explanation:

Using Linear combination method to solve:

2d + e = 8\\d - e = 4\\

Since "e" have the same coefficient in both equation with opposite operator; we will add.

(2d + d) + (e - e) = (8 + 4)\\3d = 12\\

Divide both side by coefficient of d which is 3

\frac{3d}{3}  = \frac{12}{3} \\d = 4\\

Since d = 4; put 'd' into any of the equation to get 'e'

2d + e = 8\\2(4) + e = 8\\8 + e = 8\\e = 8 - 8\\e = 0\\

Therefore, the solution is (4, 0)

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Step-by-step explanation:

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3 years ago
Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three h
dangina [55]

Answer:

Step-by-step explanation:

To solve the differential equation

dy/dx = 5x^4(1 + y²)^(3/2)

First, separate the variables

dy/(1 + y²)^(3/2) = 5x^4 dx

Now, integrate both sides

To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)

dy = (1/cos²u)du

So,

dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du

= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)

Because cos²u + sin²u = 1 (Trigonometric identity),

The equation becomes

[1/(1/cos²u)^(3/2) × 1/cos²u] du

= cos³u/cos²u

= cosu

Integral of cosu = sinu

But y = tanu

Therefore u = arctany

We then have

cos(arctany) = y/√(1 + y²)

Now, the integral of the equation

dy/(1 + y²)^(3/2) = 5x^4 dx

Is

y/√(1 + y²) = x^5 + C

So

y - (x^5 + C)√(1 + y²) = 0

is the required implicit solution

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3 years ago
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