All categories

2 years ago

-13=r/9+8 what is r equal to?

Mathematics

1 answer:

AURORKA [14]2 years ago

7 0

Answer: r=(-189)

first you subtract both sides by 8 and get -21=r/9

then you multiply 9 with each side and get -189 = r

you can check by rewriting the problem with -189 as r and solving the equation

You might be interested in

Simplify : 4y^2+12y-40\y^2+2y-8

PSYCHO15rus [73]

2(2y^4 + 7y^3 - 4y^2 -20)
___________________
Y^2

4y^2 - 40/y^2 + 14y - 8

2(-2y^4 - 7y^3 + 4y^2 + 20)
_ _____________________
Y^2

8 0

3 years ago

-5X +1 92 - 321+13 - 12X

Bogdan [553]

Answer:

Step-by-step explanation:

There’s no equal sign.

8 0

3 years ago

F(x) = e^-x . Find the equation of the tangent to f(x) at x=-1

natima [27]

Answer:

The equation of the tangent line is given by the following equation:

$\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)$

General Formulas and Concepts:

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

*Note:

Recall that the definition of the derivative is the slope of the tangent line.

Step 1: Define

Identify given.

$\displaystylef(x) = e^{-x} \\x = -1$

Step 2: Differentiate

[Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
$\displaystyle f'(x) = e^{-x}(-x)'$
[Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
$\displaystyle f'(x) = -e^{-x}(x)'$
[Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
$\displaystyle f'(x) = -e^{-x}$

Step 3: Find Tangent Slope

[Derivative] Substitute in x = 1:
$\displaystyle f'(1) = -e^{-1}$
Rewrite:
$\displaystyle f'(1) = \frac{-1}{e}$

∴ the slope of the tangent line is equal to $\displaystyle \frac{-1}{e}$ .

Step 4: Find Equation

[Function] Substitute in x = 1:
$\displaystyle f(1) = e^{-1}$
Rewrite:
$\displaystyle f(1) = \frac{1}{e}$

∴ our point is equal to $\displaystyle \bigg( 1, \frac{1}{e} \bigg)$ .

Substituting in our variables we found into the point-slope form general equation, we get our final answer of:

$\displaystyle \boxed{ y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg) }$

∴ we have our final answer.

---

Learn more about derivatives: brainly.com/question/27163229

Learn more about calculus: brainly.com/question/23558817

---

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

3 0

1 year ago

If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

 dy
Find —— for an implicit function:
 dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

 dy
Now, isolate —— in the equation above:
 dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: implicit function derivative implicit differentiation chain product rule differential integral calculus

6 0

2 years ago

Hello What is 5X+2x-3x

Sedaia [141]

I think 5X+1x. Sorry if it's wrong.

5 0

2 years ago

Read 2 more answers