Hello from MrBillDoesMath!
Answer: N = 143
Discussion:
This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers: 11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of 3 consecutive primes. I went back to my list of primes, added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on 43, 47, and 53!
43 + 47 + 53 = 143 !
Therefore N = 143. It's the sum of 43, 47, and 53 as well as the product of 11 and 13.
Thank you,
MrB
Subtract 10 from both sides.
Since the square of a number, x, equals a negative number, -1, the answer cannot be a real number because the square of a real number is always non-negative. Therefore, there is no real number solution to this equation.
Answer: <span>B. no real number solutions</span>
Answer: z + 17
Step-by-step explanation:
1) DISTRIBUTE:
11 + 2z + 6 - z
2) COMBINE LIKE TERMS
z + 17
Step-by-step explanation:
Applying rules of exponents to solve the given problems;
4^3 x 4^5 =
5^8 ÷ 5^-2 =
(6^3 ) ^ 4 =
For these problems, the applicable rules of exponents are;
aᵇ x aⁿ = aᵇ⁺ⁿ
aᵇ ÷ aⁿ = aᵇ⁻ⁿ
(aᵇ)ˣ = aᵇˣ
For the first problem; 4³ x 4⁵
aᵇ x aⁿ = aᵇ⁺ⁿ
4³ x 4⁵ = 4³⁺⁵ = 4⁸
Second problem: aᵇ ÷ aⁿ = aᵇ⁻ⁿ
5⁸ ÷ 5⁻² = 5⁸⁻⁽⁻²⁾ = 5⁸⁺² = 5¹⁰
Third problem; (aᵇ)ˣ = aᵇˣ
(6³)⁴ = 6³ˣ⁴ = 6¹²
