All categories

3 years ago

Solve for u: u/p + u/q =m, if , p≠−q

Mathematics

2 answers:

nikitadnepr [17]3 years ago

5 0

Answer:

The solution for u is:

$u = \frac{mpq}{q+p}$

Step-by-step explanation:

The first step to solve this problem is finding the least common multiplicator between p and q, that is pq, so:

$\frac{u}{p} + \frac{u}{q} = m$

$\frac{uq + up}{pq} = m$

$uq + up = mpq$

$u(q + p) = mpq$

$u = \frac{mpq}{q+p}$

katovenus [111]3 years ago

4 0

U/p + u/q = m

uq/pq + up/pq = m

u^2pq/pq =m

u^2 = m
u = sqrt(m)

You might be interested in

Prove that P (P) = (QA ~ Q)] is a tautology.

alekssr [168]

Answer:

The statement $P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)]$ is a tautology.

Step-by-step explanation:

A tautology is a formula which is "always true" that is, it is true for every assignment of truth values to its simple components.

To show that this statement is a tautology we are going to use a table of logical equivalences:

$P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)] \equiv$

$\equiv (P \land [(\lnot P)\rightarrow (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [(\lnot P)\rightarrow (Q \land \lnot Q)])$ by the logical equivalences involving bi-conditional statements

$\equiv (P \land [\lnot(\lnot P)\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [\lnot(\lnot P)\lor (Q \land \lnot Q)])$ by the logical equivalences involving conditional statements

$\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [ P\lor (Q \land \lnot Q)])$ by the Double negation law

$\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))$ by De Morgan's law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))$ by the Negation law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor \lnot(\lnot Q))$ by De Morgan's law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Double negation law

$\equiv (P \land P) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Identity law

$\equiv (P) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Idempotent law

$\equiv (P) \lor(\lnot P \land \lnot P\land (Q\lor \lnot Q))$ by the Commutative law

$\equiv (P) \lor(\lnot P \land \lnot P\land T)$ by the Negation law

$\equiv (P) \lor(\lnot (P \lor P)\land T)$ by De Morgan's law

$\equiv (P) \lor(\lnot (P)\land T)$ by the Idempotent law

$\equiv (P \lor\lnot P) \land(P \lor T)$ by the Distributive law

$\equiv (T) \land(P \lor T)$ by the Negation law

$\equiv (T) \land(T)$ by the Domination law

$\equiv T$

8 0

4 years ago

Plz answer ...Find f(-3).

Cloud [144]

A, -5 because you can use synthetic division

6 0

3 years ago

Read 2 more answers

Find the third side in simplest radical form: 6 √90

Rus_ich [418]

Your answer will be 96 hope this helps :)

6 0

3 years ago

Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

3 years ago

The cones below are similar, although not drawn to scale.

Dmitry [639]

9 ft
Set up a ratio...6/18=x/27...reduce to 1/3=x/27...cross multiply and solve for x.
3x=27 so x=9

5 0

3 years ago

Read 2 more answers