Answer:
6, 8, 10.
Step-by-step explanation:
Let n be any integer.
Then our first even integer will be
And the consecutive integers will be and
We want to find the integers such that the sum of <em>twice </em>the smallest number (2n) and <em>three times</em> the largest number (2n+4) is 42. In other words:
Solve for n. Distribute:
Combine like terms:
Subtract 12 from both sides:
Divide both sides by 10:
So, the value of n is 3.
This means that our first even integer is 2(3) or 6.
So, our sequence of integers is: 6, 8, 10.
And we're done!
Notes:
We use 2n because anything integer multiplied by 2 ensures that the resulting number is even. Starting out, n can be either even or odd, but by multiplying it by 2, we <em>will </em>get an even number.
A. coefficient is the answer
I'm not taking you through the method of solution.
You can do it with two numbers: 21.6619 and -1.6619 (both rounded) .
Answer:
see explanation
Step-by-step explanation:
The vertex of f(x) = x² is at (0, 0)
To find the vertex of g(x) = x² - 8x + 7 ← in standard form
The x- coordinate of the vertex is
= -
where a = 1 and b = - 8
= - = 4
substitute x = 4 into the equation for corresponding y- coordinate
y = 4² - 8(4) + 7 = 16 - 32 + 7 = - 9
The vertex of g(x) is at (4, - 9)
thus (0, 0) → (4, - 9) is 4 units right and 9 units down → option C
Answer:
Add 12+4
Step-by-step explanation: