Solve -Q=1/2p+15,for p

Mathematics

2 answers:

alukav5142 [94]3 years ago

4 0

P=-2Q-30 is the answer

3241004551 [841]3 years ago

4 0

Hello there!

-q = 1/2p + 15
Solve for P
-2pq = 30p + 1
-2pq - 30p= 1
Factor out the variable P
p(-2q - 30)=1
Now we can divide both sides by -2q-30
p(-2q - 30)/-2q - 30 = 1/-2q - 30
P = -1/(2q+30)

I hope I helped!

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3 years ago

What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .