Answer:
Step-by-step explanation:
Method 1: Taking the log of both sides...
So take the log of both sides...
5^(2x + 1) = 25
log 5^(2x + 1) = log 25 <-- use property: log (a^x) = x log a...
(2x + 1)log 5 = log 25 <-- distribute log 5 inside the brackets...
(2x)log 5 + log 5 = log 25 <-- subtract log 5 both sides of the equation...
(2x)log 5 + log 5 - log 5 = log 25 - log 5
(2x)log 5 = log (25/5) <-- use property: log a - log b = log (a/b)
(2x)log 5 = log 5 <-- divide both sides by log 5
(2x)log 5 / log 5 = log 5 / log 5 <--- this equals 1..
2x = 1
x=1/2
Method 2
5^(2x+1)=5^2
2x+1=2
2x=1
x=1/2
Answer:
2 x e-4
Step-by-step explanation:
7) You have the opposite. You need the adjacent.
x = 7.79289... =
<h2>7.8</h2><h2 />
8) You have the hypotenuse. You need the Opposite.
x = sin37 × 13
x = 7.8235... =
<h2>7.8</h2><h2 />
9) You have the hypotenuse. You need the opposite.
x = sin59 × 11
x = 9.4288.... =
<h2>9.4</h2><h2 />
10) You have the hypotenuse. You need the opposite.
x = sin53 × 11
x = 8.7459.... =
<h2>8.7</h2>
Answer:
B
Step-by-step explanation:
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>
