d = 3 , a₁₂ = 40 and S
= 7775
In an arithmetic sequence the nth term and sum to n terms are
<h3>• a
= a₁ + (n-1)d</h3><h3>• S
=
[2a + (n-1)d]</h3><h3>
where d is the common difference</h3><h3>a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3</h3><h3>a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40</h3><h3>S₁₀₀ =
[(2×7) +(99×3)</h3><h3> = 25(14 + 297) = 25(311)= 7775</h3>
Answer:
x=3
Step-by-step explanation:
first let's try to get just 3x on one side. to do this we can subtract 2 from both side of the equation.
3x+(2-2)=11-2
3x=9
now we have to isolate the variable we can divide both sides by 3:
3x/3=9/3
x=3
and you get the answer that x is 3
Answer:
10.15
Step-by-step explanation:
Subtract:
10.15 - 10.1 = 0.05
10.15 is greater than 10.1 by 0.05
~
Answer: x=1.5811389999999999 , or 1.59
Step-by-step explanation:
First, Subtract 20 from both sides.
12x2+20−20=50−20
12x2=30
Then, Divide both sides by 12:
12x^2/12 = 30/12
⇒ you get : x^2 = 5/2
After, Take square root.
x= ± √ 5/2
Finally you answer is going to be =1.5811389999999999 OR, 1.59
* Hopefully this helps:) Mark me the brainliest:)
<em>∞ 234483279c20∞</em>
