Find a set of four consecutive positive integers such that the greatest integer in the set is twice the least integer in the set
1 answer:
Answer:
four consecutive integers are
3,4,5,6
Step-by-step explanation:
Let x, x+1, x+2 and x+3 are the four consecutive positive integers
greatest integer is x+3
least integer is x
the greatest integer in the set is twice the least integer
subtract x from both sides
the value of x is 3
x+1 is 4
x+2 is 5
x+3 is 6
so four consecutive integers are
3,4,5,6
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Answer:
The value of n is -6
Step-by-step explanation:
- If the function f(x) is translated k units up, then its image is g(x) = f(x) + k
- If the function f(x) is translated k units down, then its image is g(x) = f(x) - k
- The vertex form of the quadratic function is f(x) = a(x - h)² + k, where a is the coefficient of x² and (h, k) is the vertex
∵ k(x) = x²
→ Its graph is a parabola with vertex (0, 0)
∴ The vertex of the prabola which represents it is (0, 0)
∵ The given graph is the graph of p(x)
∵ Its vertex is (0, -6)
∴ h = 0 and k = -6
∵ a = 1
→ Substitute them in the form above
∴ p(x) = 1(x - 0)² + -6
∴ p(x) = x² - 6
→ Substitute x² by k(x)
∴ p(x) = k(x) - 6
∵ p(x) = k(x) + n
→ By comparing the two right sides
∴ n = -6
∴ The value of n is -6
Look at the attached figure for more understanding
The red parabola represents k(x)
The blue parabola represents p(x)
