Can someone please help??

What number is shown on the graph ?

Natalka [10]

El numero daaaaaaaaa

4 0

3 years ago

Plz help ASAP I will add branliest to the first answer, I need to solve for the letter

harkovskaia [24]

Answer:

6. r=2

8.x=3

Step-by-step explanation:

6. 4r-3=5

4r-3=5 add 3 to both sides

4r=8 divide 4 by both sides

r=2

8. 5x-6=9

5x-6=9 add 6 to both sides

5x=15 divide 5 by both sides

x=3

6 0

2 years ago

Read 2 more answers

What is 29/4 simplyfied?

Mars2501 [29]

Answer:

7 and 1/4 or 7.25

Step-by-step explanation:

28/4 = 7

1/4 = 0.25

7 + 0.25 = 7.25

5 0

3 years ago

Read 2 more answers

Let f(x)=x^3-x-1

alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

6 0

3 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago