Solve the following systems using substitution: x+y=4 x^2+y^2=9
1 answer:
You might be interested in
I believe you meant to write
?
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) <span>≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 </span><span>≤ 5
Add 4 both sides:
x^2 </span>≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x <span>≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.</span>
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) <span>≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 </span><span>≤ 5
Add 4 both sides:
x^2 </span>≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x <span>≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.</span>
Answer:
Step-by-step explanation:
If Quan had 15 and Kaden had 3 times as much, then Kaden had 15(3) = 45. If he spent 10 then he had 35 left. The amount of money Quan had after his mom gave him some was 4/5 the amount of Kaden's 35 (let x = the amount of money Quan has after his mom gives him some):
35 divides by the 5 in the denominator 7 times, and 7 times 4 is 28. That's how you get your 28 in part a.
For part b., he started with 15 and ended with 28, so his mom gave him 28-15=13. That's how you get the 13.