Fill in the blanks 3/4 (12x-16) = 1/3 (?x-?)

Mathematics

1 answer:

Marina CMI [18]2 years ago

3 0

Answer:

Step-by-step explanation:

3/4 * 12 = 9

3/4 * 16 = 12

So rewriting the left side, you get

9x - 12

The numbers written on either side of the x when multiplied by 1/3 must equal 9 and 12

9*3 = 27

12*3 = 36

The answer is

1/3 ( 27x - 36)

You might be interested in

4x-6y=17 4x+2y=21 solua

den301095 [7]

Subtracting 4x+2y=21 from 4x-6y=17 will cancel out the x terms and result in -8y=-4. Then, dividing both sides by -8 results in y=1/2. In order to find x, plug 1/2 for y in either equation. If we use the first equation, the new equation looks like: 4x-6(1/2)=17. Multiplying -6 and 1/2 will simplify the equation to 4x-3=17. Then, adding three to both sides results in 4x=20. The last step is to divide both sides by 4 in order to get x by itself. x=5. So, the coordinate point that satisfies both equations is (5,1/2).

6 0

1 year ago

How many applications of integration by parts are required to evaluate integral (x^7)(e^x) dx?

charle [14.2K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2860229

_______________

Make it more general, and see what happens when you try to reduce the exponent of x in the following integral:

$\mathsf{\mathtt{I}_0=\displaystyle\int\! x^k\cdot e^x\,dx\qquad\qquad (k\ge 1,~~k\in\mathbb{N})}$

Now, integrate it by parts:

$\begin{array}{lcl}
\mathsf{u=x^k}&\quad\Rightarrow\quad&\mathsf{du=k\cdot x^{k-1}\,dx}\\\\
\mathsf{dv=e^x\,dx}&\quad\Leftarrow\quad&\mathsf{v=e^x}
\end{array}$

$\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\! v\,du}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-\int\! e^x\cdot k\cdot x^{k-1}\,dx}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\int\! x^{k-1}\cdot e^x\,dx}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\cdot \mathtt{I}_1}$

where $\mathsf{\mathtt{I}_1=\displaystyle\int\! x^{k-1}\cdot e^x\,dx.}$

So after one iteration, the exponent of x was decreased by one unit.

The question is: after how many iterations will the exponent of x equals zero?

After exactly k iterations, of course.

Therefore, for k = 7, you have to apply integration by parts 7 times, to get rid of that polynomial factor. Then, there will be one last integral left to evaluate:

$\mathsf{\displaystyle\int\! e^x\,dx}$

But this one doesn't need to be evaluated by parts. You can directly write the result:

$\mathsf{\displaystyle\int\! e^x\,dx=e^x+C}$

Shortly, for the integral

$\mathsf{\mathtt{I}_0=\displaystyle\int\! x^7\cdot e^x\,dx}$

you have to apply integration by parts 7 times (not 8 times).

I hope this helps. =)

Tags: <em>indefinite integral integration by parts reduction formula product polynomial exponential differential integral calculus</em>

8 0

3 years ago

What is the y-intercept of the line passing through the point (9-7) and has a slope of -1/8

Lina20 [59]

Y - y1 = m(x - x1)
y + 7 = (-1/8)(x - 9)
y + 7 = (-1/8)x + 1.125
Subtract 7 from both sides
y = -1/8x - 5.875

4 0

2 years ago

Help. again............................

Georgia [21]

What do u need help with ????

7 0

2 years ago