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Romashka-Z-Leto [24]

2 years ago

5

14 - 2(11 - 5) +

ss="latex-formula">=

1 answer:

Elodia [21]2 years ago

6 0

Answer:

51

Step-by-step explanation:

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How many terms are in the algebraic expression 11;; +3 ?

xxTIMURxx [149]

That would be two terms

4 0

2 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago

What is the solution of √x-5 = x-11

PIT_PIT [208]

Hey!

To solve x in this equation we must first add five to both sides to get $\sqrt{x}$ on its own.

<em>Original Equation :</em>
$\sqrt{x} - 5 = x - 11$

<em>New Equation {Added 5 to Both Sides} :</em>
$\sqrt{x} =x-6$

Now we must square both sides of the equation.

<em>Old Equation :</em>
$\sqrt{x} =x-6$

<em>New Equation {Changed by Squaring Both Sides} :</em>
$x= x^{2} -12x+36$

And now we must solve the new equation.

Step 1 - Switch sides
$x^{2} -12x+36=x$

Step 2 - Subtract x from both sides
$x^{2} -12x+36-x=x-x$

Step 3 - Simplify
$x^{2} -13x+36=0$

Now we need to solve the rest of the equation using the quadratic formula.

$\frac{-(-13)+ \sqrt{(-13) ^{2}-4*1*36 } }{2*1}$

${13+ \sqrt{(-13)^{2}-4*1*36 }=13+ \sqrt{25}$

$\frac{13+ \sqrt{25}}{2}$

$\sqrt{25}=5$

$\frac{13+5}{2}$

$\frac{18}{2}$

9

$\frac{-(-13)- \sqrt{(-13) ^{2} -4*1*36} }{2*1}$

4

<em>So, this means that in the equation $\sqrt{x} -5=x-11$ ,</em> x = 9 <em>and </em>x = 4.

Hope this helps!

- Lindsey Frazier ♥

4 0

2 years ago

HELP ME PLZ ASAP!!!!!!!!!!!

kondor19780726 [428]

(8q)/4 ..................

3 0

3 years ago

6(22+9) what is this answer

scoundrel [369]

If you would like to solve 6 * (22 + 9), you can calculate this using the following steps:

6 * (22 + 9) = 6 * 31 = 186

The correct result would be 186.

5 0

3 years ago

Read 2 more answers