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3 years ago

Plz help for last 2?

Mathematics

1 answer:

VikaD [51]3 years ago

8 0

139.25 in = 11 ft 7.25 in

149 3/16 in= 12 ft 5.2 in

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If dy/dx = x(y-1), find the equation of the tangent line at x=-2 with f(-2)=3

Lapatulllka [165]

Answer:

The equation of the tangent line passing through the point (-2,3) is

4 x + y +5 =0

Step-by-step explanation:

Step(i):-

Given that the slope of the tangent

$m = \frac{dy}{dx} = x(y-1)$

m = -2( 3-1) = -4

Given point x = -2

y = f(-2) =3

∴The given point ( x₁ , y₁) = ( -2 ,3)

Step(ii):-

The equation of the tangent line passing through the point (-2,3)

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

$y -3 = -4(x-(-2))$

y -3 = -4( x+2)

y-3 = -4x -8

4x + y -3+8=0

4x +y +5=0

Step(iii):-

The equation of the tangent line passing through the point (-2,3) is

4 x + y +5 =0

3 0

3 years ago

How do you solve x/2=2/5

Anarel [89]

You multiply 2 and 2, and then divide it by 5

3 0

3 years ago

Eeeek help me please

Talja [164]

With what? There’s no picture

8 0

3 years ago

What would the constant rate of change be?

deff fn [24]

Answer:

- 8

Step-by-step explanation:

An equation in the form y = mx + c ( m is the slope and c is the y- intercept )

Has a constant rate of change = m

given

$a_{n}$ = 12 + (n - 1)(- 8) ← distribute and simplify

y = 12 - 8n + 8

y = - 8n + 20 ← in the form y = mx + c

with constant rate of change = - 8

4 0

4 years ago

The revenue equation (in hundreds of millions of dollars) for barley production in a certain country is approximated by R(x) = 0

STatiana [176]

Answer:

The marginal-revenue equation is $R'(x)=0.1226x+1.1584$ .

The marginal revenue for the production of 200,000,000 bushels is $R'(x)=1.4036$ .

The marginal revenue for the production of 650,000,000 bushels is $R'(x)=9.1274$ .

Step-by-step explanation:

The derivative $R'(x)$ of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.

We know the revenue function, $R(x) = 0.0613x^{2} +1.1584x+2.4813$ . Therefore, the marginal revenue function is

$R'(x)=\frac{d}{dx}R(x) = \frac{d}{dx}(0.0613x^{2} +1.1584x+2.4813)\\\\=\frac{d}{dx}\left(0.0613x^2\right)+\frac{d}{dx}\left(1.1584x\right)+\frac{d}{dx}\left(2.4813\right)\\\\R'(x)=0.1226x+1.1584$

To find the marginal revenue for the production of:

200,000,000 bushels we use x = 2, since x is in units of hundreds of million of dollars.

$R'(x)=0.1226\cdot \:2+1.1584\\\\R'(x)=0.2452+1.1584\\\\R'(x)=1.4036$

650,000,000 bushels we use x = 65, since x is in units of hundreds of million of dollars.

$R'(x)=0.1226\cdot \:65+1.1584\\\\R'(x)=7.969+1.1584\\\\R'(x)=9.1274$

3 0

3 years ago