side length of square =4
side length of cube=3
hence the correct statement is,
the side length of the cube is less than the side length of the square
Answer: the function g(x) has the smallest minimum y-value.
Explanation:
1) The function f(x) = 3x² + 12x + 16 is a parabola.
The vertex of the parabola is the minimum or maximum on the parabola.
If the parabola open down then the vertex is a maximum, and if the parabola open upward the vertex is a minimum.
The sign of the coefficient of the quadratic term tells whether the parabola opens upward or downward.
When such coefficient is positive, the parabola opens upward (so it has a minimum); when the coefficient is negative the parabola opens downward (so it has a maximum).
Here the coefficient is positive (3), which tells that the vertex of the parabola is a miimum.
Then, finding the minimum value of the function is done by finding the vertex.
I will change the form of the function to the vertex form by completing squares:
Given: 3x² + 12x + 16
Group: (3x² + 12x) + 16
Common factor: 3 [x² + 4x ] + 16
Complete squares: 3[ ( x² + 4x + 4) - 4] + 16
Factor the trinomial: 3 [(x + 2)² - 4] + 16
Distributive property: 3 (x + 2)² - 12 + 16
Combine like terms: 3 (x + 2)² + 4
That is the vertex form: A(x - h)² + k, whch means that the vertex is (h,k) = (-2, 4).
Then the minimum value is 4 (when x = - 2).
2) The othe function is <span>g(x)= 2 *sin(x-pi)
</span>
The sine function goes from -1 to + 1, so the minimum value of sin(x - pi) is - 1.
When you multiply by 2, you just increased the amplitude of the function and obtain the new minimum value is 2 (-1) = - 2
Comparing the two minima, you have 4 vs - 2, and so the function g(x) has the smallest minimum y-value.
To model this situation, we are going to use the compound interest formula:
where
is the final amount after
years
is the initial deposit
is the interest rate in decimal form
is the number of times the interest is compounded per year
is the time in years
For account A:
We know for our problem that
and
. Since the interest is compounded monthly, it is compounded 12 times per year; therefore,
. Lets replace those values in our formula:
For account B:
,
,
. Lest replace those values in our formula:
Since we want to find the time,
, <span>when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>
Now that we have our equation, we just need to solve for
:
![2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000](https://tex.z-dn.net/?f=2000%5B%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D%2B%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D%5D%3D5000%20)
![t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})](https://tex.z-dn.net/?f=t%5B12ln%281.001875%29%2B12ln%281.0025%20%29%5D%3Dln%28%20%5Cfrac%7B5%7D%20%7B2%7D%29)
We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
