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2 years ago

Combine like-terms of 6x-2x-2=-3x+4(x+7)

Mathematics

2 answers:

den301095 [7]2 years ago

7 0

Step-by-step explanation:

The answer is x = 10

Leya [2.2K]2 years ago

3 0

3x-2 = 3x+4x+28x

3x-2 = 7x + 28

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Select all the roots of the following equation.

strojnjashka [21]

Answer:

5, -9

Step-by-step explanation:

(x - 5)(x + 9) = 0 These are factors, to get the roots, just set each one equal to zero

x - 5 = 0

x = 5

x + 9 = 0

x = -9

6 0

2 years ago

Read 2 more answers

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago

HELP FOR BRAINLESS !!!!

garik1379 [7]

Answer:

the blue one

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find the distance traveled by a particle with position (x, y as t varies in the given time interval. x = 3?sin2 t, y = 3?cos2 t,

ANTONII [103]

Assuming the particle's position is given by

$\begin{cases}x(t)=3-\sin2t\\y(t)=3-\cos2t\\0\le t\le4\end{cases}$

then the distance traveled over the interval is

$\displaystyle\int_{t=0}^{t=4}\sqrt{\left(\dfrac{\mathrm dx(t)}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy(t)}{\mathrm dt}\right)^2}\,\mathrm dt$
$=\displaystyle\int_0^4\sqrt{(-2\cos2t)^2+(2\sin2t)^2}\,\mathrm dt$
$=\displaystyle\int_0^4\sqrt{4\cos^22t+4\sin^22t}\,\mathrm dt$
$=\displaystyle2\int_0^4\mathrm dt$
$=2(4-0)=8$

5 0

3 years ago

Find the volume of the solid. The volume of the solid is ____ cubic centimeters

Illusion [34]

Is there a picture ?

6 0

2 years ago