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denis-greek [22]

3 years ago

5

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

rameter. x = 5 + ln(t), y = t2 + 1, (5, 2)

1 answer:

liubo4ka [24]3 years ago

4 0

Im sorry i need points , i am really sorry

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What is the slope of the line perpendicular to a line that contains the points (-5,4) and (-2,4)

steposvetlana [31]

Answer:

the slope of the line is: 0

Step-by-step explanation:

4-4=0

-2-5= 3

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A system of equations has 1 solution. If 4x - y = 5 is one of the equations, which could be the other equation?

Lera25 [3.4K]

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3 years ago

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ehidna [41]

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3 years ago

Which equation represents the line that passes through points (1, –5) and (3, –17)? y = negative 6 x + 1 y = 6 x + 1 y = negativ

tia_tia [17]

Answer:

[see below]

Step-by-step explanation:

Finding the slope of the two points:

$m=\frac{-17+5}{3-1}=\frac{-12}{2}=\boxed{-6}$

The slope is -6.

Using an incomplete slope-intercept form equation to find the y-intercept:

$y=6x+b\\\\-5=6(1)+b\\\\-5=6+b\\\\-5-6=6-6+b\\\\\boxed{1=b}$

The equation for the line passing the points is $y=-6x+1$ .

The equation in standard from would be $6x+y=1$ .

4 0

3 years ago