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>
The answer is B. To find the Circumference (what you're looking for) you use 2×pi×radius. Which is 2×pi×7. In terms of pi that's just 2×7, which equals 14. So that means your answer is 14pi.
The set of two-digit primes is {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
Of that list, the following primes are mirror images of each other
13 and 31
17 and 71
37 and 73
79 and 97
Note: we ignore 11 since 11 flips to 11 which is not distinct from its original
If you're looking for the largest prime of this form, then its 97
If you're looking for the largest gap, then subtract each pair
31-13 = 18
71-17 = 54
73-37 = 36
97-79 = 18
We see that 71 and 17 have the largest gap
Answer:
The cost of
A small bucket = x = $8
A Larger bucket = y = $13
Step-by-step explanation:
Let the cost of
A small bucket = x
A Larger bucket = y
Kim sold 3 small buckets and 14 large buckets of popcorn for a total of $206.
3x + 14y = 206... Equation 1
Melody sold 11 small buckets and 11 large buckets of popcorn for a total of $231.
11x + 11y = 231.... Equation 2
What is the cost for a small bucket and a large bucket of popcorn?
3x + 14y = 206... Equation 1
11x + 11y = 231 ....... Equation 2
We solve using Elimination method
Multiply Equation 1 by 11 and Equation 2 by 3 to Eliminate x
33x + 154y = 2266.... Equation 3
33x + 33y = 693....... Equation 4
We subtract Equation 4 from Equation 3
121y = 1573
y = 1573/121
y = $13
Solving for x
3x + 14y = 206... Equation 1
3x + 14 × 13 = 206
3x + 182 = 206
3x = 206 - 182
3x = 24
x = 24/3
x = $8
Hence, the cost of
A small bucket = x = $8
A Larger bucket = y = $13
