Answer:
first one is correct................
This solution is arrived at using the principles of Quadratic Functions and Inequalities. It is to be noted that the function increases at (-8, 4).
<h3>What are Quadratic Functions?</h3>
A quadratic function can be defined as that is written in the following way: f(x) = ax2 + bx + c: Where a, b, and c are numbers not equal to zero.
<h3>How do we solve the problem above?</h3>
To solve the problem f(4x - 3) ≥ f(2 - x^2), Df = (-8 , 4), we:
- Rewrite
- Collect like terms
- Expand
- Factorize, then
- Solve for <em>x</em>.
A) 4x - 3 ≥ 2 - x²
Rewrite as: x² + 4x - 2 - 3 ≥ 0
B) Collect the like terms
x² + 4x - 5 ≥ 0
C) Expand to have x² + 5x - x - 5 ≥ 0
D) Factorizing the expression we have
x(x + 5) - 1(x + 5) ≥ 0, from this we Factor away x + 5
(x - 1)(x + 5) ≥ 0
E) Solving for x we obtain: x ≥ 1 or x ≥ -5
This can also be rewritten to read:
-5 ≤ x ≤ 1
Learn more about quadratics and inequalities at:
brainly.com/question/2237134
The answer is D to the next step
We know the equation for company A is P(t) = 1.8(1.4)ᵗ.
and what we know about company B is that their equation is linear, and we also have 3 months in a table of values, and we can simply use two points from there to get the equation of company B, let's do so using (3,5) and (7, 25),
![\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 3 &,& 5~) % (c,d) &&(~ 7 &,& 25~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{25-5}{7-3}\implies \cfrac{20}{4}\implies 5 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-5=5(x-3) \\\\\\ y-5=5x-15\implies y=5x-10\implies p(t)=5t-10](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%203%20%26%2C%26%205~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%207%20%26%2C%26%2025~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%20%0A%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B25-5%7D%7B7-3%7D%5Cimplies%20%5Ccfrac%7B20%7D%7B4%7D%5Cimplies%205%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-5%3D5%28x-3%29%0A%5C%5C%5C%5C%5C%5C%0Ay-5%3D5x-15%5Cimplies%20y%3D5x-10%5Cimplies%20p%28t%29%3D5t-10)
now, the end of the year will be at the twelfth month, namely t = 12, and the four month well we know that t = 4.
how did each one fare?
![\bf \stackrel{\textit{at the 4th month}}{\stackrel{\textit{company A}}{P(4)=6.91488}\qquad \qquad \stackrel{\textit{company B}}{p(4)=10}} \\\\\\ \stackrel{\textit{at the end of the year}}{\stackrel{\textit{company A}}{P(12)\approx 102.04904}\qquad \qquad \stackrel{\textit{company B}}{p(12)=70}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bat%20the%204th%20month%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bcompany%20A%7D%7D%7BP%284%29%3D6.91488%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bcompany%20B%7D%7D%7Bp%284%29%3D10%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7B%5Ctextit%7Bat%20the%20end%20of%20the%20year%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bcompany%20A%7D%7D%7BP%2812%29%5Capprox%20102.04904%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bcompany%20B%7D%7D%7Bp%2812%29%3D70%7D%7D)
so, notice, on the fourth month B did better than A, but at the end of the year, A ended up better off than B.