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3 years ago

Suppose that a point moves along some unknown curve y = f(x) in such a way that, at

Mathematics

1 answer:

spayn [35]3 years ago

8 0

Answer:

f(x) = x³/3 - 5/3

Step-by-step explanation:

To get the equation of the curve, integrate the slope

f(x) = integral of x²

= (x^(2+1))/3 + c

f(x) = x³/3 + c

To find c, use the given point

1 = 2³/3 + c

c = 1 - 8/3 = -5/3

f(x) = x³/3 - 5/3

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Brrunno [24]

Y=-1/2x+5 slope=-1/2 because it goes down one unit and to the right 2 units, and the y intercept is 5 because that's where the line intercepts the y-axis!

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2 years ago

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Please help I Think I'm wrong!!!!

Allisa [31]

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3 years ago

A line intersects the points (-5, 1) and (-2,7). What is the slope of the line in simplest form? m=[?}

IrinaK [193]

Step-by-step explanation:

To find slope : y2- y1 / x2 - x1

we have ( -5 , 1 ) which is ( x1 , y1 )

and ( -2 , 7 ) is ( x2 , y2 )

so; 7 - 1 / -2 - (-5)

6/3

= <u>2</u>

7 0

2 years ago

The population of a city is 250,000 and the annual growth rate is 2.2%. Write an equation to model the population y after x year

Zinaida [17]

Answer:

$y= 250000(1.022)^x$

Step-by-step explanation:

The population of a city is 250,000 and the annual growth rate is 2.2%

General equation for exponential growth is

$y= P(1+r)^x$

Where y is final population and P is the initial population

'r' is the rate of growth and x is the number of years

p = 25000 and r= 2.2%= 0.022. Replace all the values in the general equaiton

$y= P(1+r)^x$

$y= 250000(1.022)^x$

3 0

3 years ago

An equation parallel and perpendicular to 4x+5y=19

UNO [17]

Answer:

Parallel line:

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

$y=\frac{5}{4}x-\frac{1}{2}$

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

$4x+5y=19$

we can change it into y=mx+b form

$5y=-4x+19$

$y=-\frac{4}{5}x+\frac{19}{5}$

so,

$m=-\frac{4}{5}$

Parallel line:

we know that slope of two parallel lines are always same

so,

$m'=-\frac{4}{5}$

Let's assume parallel line passes through (1,1)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-1=-\frac{4}{5}(x-1)$

now, we can solve for y

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

$m'=\frac{5}{4}$

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-2=\frac{5}{4}(x-2)$

now, we can solve for y

$y=\frac{5}{4}x-\frac{1}{2}$

4 0

3 years ago