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3 years ago

What is the answer to 2x + 9(x – 1) = 8(2x + 2) – 5

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

-4 =x

Step-by-step explanation:

2x + 9(x – 1) = 8(2x + 2) – 5

Distribute

2x +9x -9 = 16x+16 -5

Combine like terms

11x-9 = 16x +11

Subtract 11x from each side

11x-9-11x = 16x-11x+11

-9 = 5x+11

Subtract 11 from each side

-9-11 = 5x+11-11

-20 = 5x

Divide by 5

-20/5 = 5x/5

-4 =x

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y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the

vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

$y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},$

where $c_1, c_2 \in \mathbb{R}$ . We now want to find a solution $y$ such that $y(-1)=3$ and $y'(-1)=-3$ . Therefore, all we need to do is find the constants $c_1$ and $c_2$ that satisfy the initial conditions. For the first condition, we have: $y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.$

For the second condition, we need to find the derivative $y'$ first. In this case, we have:

$y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.$

Therefore:

$y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.$

This means that we must solve the following system of equations:

$\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.$

If we add the equations above, we get:

$\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.$

If we now substitute $c_1 = 0$ into either of the equations in the system, we get:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

$\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.$

Indeed, we can see that:

$y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3$

$y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,$

which do correspond to the desired initial conditions.

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3 years ago

What is the oblique asymptote of the function f(x) = the quantity of x squared plus x minus 2, all over x plus 1

Korvikt [17]

Hello,

y=ax+b is the asymptote
with
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3 years ago

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prohojiy [21]

Answer:

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4 years ago

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denis-greek [22]

5x3-x+ 9x7 +4 +3x11

Answer:

115-X

Explanation:

First you multiply the 5x3

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then the 9x7

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Now once you have completed the multiplication step you move on to calculating the sum. (Adding all the positive numbers)

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3 years ago

Read 2 more answers

Consider this equation 3x+ 1=2x-5

Otrada [13]

Answer: x=-6

Step-by-step explanation:

Add 5 to both sides to get 2x by itself

Yes, subtract 1 from both sides to get 3x by itself so you can subtract 2x as to not get a negative x value

One common mistake would be to accidentaly adding or subtracting 1 or 5 when they are supposed to be added/subtracted. To avoid this just go back and check your awnser by plugging it in.

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3 years ago