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>
7x20=140. 7x2000=14000. The only thing that changes the product of these numbers is the amount of zeros behind the 2. Since the only two numbers that affect the answer is the 7 and the 2. (7x2=14). The number of zeros behind the 2 affect how many zeros will be included in the product.
16x³ - 54
= 2(8x³ - 27)
= 2(2x - 3)(4x² + 6x + 9) ← answer
Answer:
All real numbers
Let's find the critical points of the inequality.
−4x+7=17
−4x+7+−7=17+−7(Add -7 to both sides)
−4x=10
−4x−1=10−1
(Divide both sides by -1)
4x=−10
4x=−10(Solve Exponent)
log(4x)=log(−10)(Take log of both sides)
x*(log(4))=log(−10)
x=log(−10)log(4)
x=NaN
