Make t the subject q= 1+3t/2-t

Mathematics

2 answers:

jok3333 [9.3K]2 years ago

5 0

Answer:

t = dfrac-1 + 2 q3 + q. Solve the equations: t = dfrac-1 + 2 q3 + q Answer: t = dfrac-1 + 2 q3 + q.

Step-by-step explanation:

Alinara [238K]2 years ago

3 0

Answer:

t = dfrac-1 + 2 q3 + q.

Step-by-step explanation:

You might be interested in

Decrease £189 by 38%

Trava [24]

I believe it would be around 117.18

7 0

3 years ago

Find the first three terms of the sequence below. T n = n 2 + 3 n + 4

madam [21]

Answer:

9,14,19

Step-by-step explanation:

You seem to be giving the expression 2n+3n+4.

Now simplify

An = 5n+4

Fill in n

5(1) +4

5(2) +4

5(3)+4

After solving and simplifying you see 9, 14 and 19 are the first 3 terms.

6 0

3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$