Let <em>f(x)</em> = <em>x</em>³ + <em>x</em> - 5. <em>f(x)</em> is a polynomial so it's continuous everywhere on its domain (all real numbers). Since
<em>f</em> (1) = 1³ + 1 - 5 = -3 < 0
and
<em>f</em> (2) = 2³ + 2 - 5 = 5 > 0
it follows by the intermediate value theorem that there at least one number <em>x</em> = <em>c</em> between 1 and 2 for which <em>f(c)</em> = 0.
Answer: 1/5 or 0.2
Step-by-step explanation:
STEP 1⇒1 1/5 = 1 · 5 + 1/5 = 5 + 1/5 = 6/5
STEP 2⇒6/5 - 2/5 = 6 - 2/5 = 4/5
STEP 3⇒4/5(-3/4) = 4 ·(-3)/5 · 4 = -12/20 = -3/4
STEP 4⇒-3/5 · 3 = -3 · 3/5 · 1 = -9/5
STEP 5⇒-9/5 ÷ (-9) = -9 · 1/5 · 9 = 9/45 = 1/5
Answer 420.
a, is the first term. d is difference.
Since this is an arithmetic progression, I use this formula.
Un = nth term
Sn = sum of terms.
First find the number of terms, which is 20.
Then use the second formula to find the sum.