What is the answer to this math question -5+1 2/3

2x - 3y = 13 x + 2y = -4

-Dominant- [34]

Answer:

(2, - 3 )

Step-by-step explanation:

Given the 2 equations

2x - 3y = 13 → (1)

x + 2y = - 4 → (2)

Rearrange (2) expressing x in terms of y by subtracting 2y from both sides

x = - 4 - 2y → (3)

Substitute x = - 4 - 2y into (1)

2(- 4 - 2y) - 3y = 13 ← distribute and simplify left side

- 8 - 4y - 3y = 13

- 8 - 7y = 13 ( add 8 to both sides )

- 7y = 21 ( divide both sides by - 7 )

y = - 3

Substitute y = - 3 into (3) for corresponding value of x

x = - 4 - 2(- 3) = - 4 + 6 = 2

Solution is (2, - 3 )

6 0

3 years ago

Hi, am I supposed to find the stationary points of the f(x) and g(x) then use the distance formula to solve for the final answer

nikdorinn [45]

9514 1404 393

Answer:

maximum difference is 38 at x = -3

Step-by-step explanation:

This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.

__

Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.

h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2

Then the derivative is ...

h'(x) = 3x² +4x -15 = (x +3)(3x -5)

This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.

The maximum difference between f(x) and g(x) is 38 at x = -3.

4 0

2 years ago

Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at <em>x</em> = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$