Home / Digital SAT Math: systems of Non linear equations in two variables-Practice Questions

Digital SAT Math: systems of Non linear equations in two variables-Practice Questions

SAT MAth Practice questions – all topics

  • Advanced Math Weightage: 35% Questions: 13-15
    • Equivalent expressions
    • Nonlinear equations in one variable and systems of equations in two variables
    • Nonlinear functions

SAT MAth and English  – full syllabus practice tests

Questions  Easy

What is the y-coordinate of the y-intercept of the graph of \(y=3^{x}+9\) ?

▶️Answer/Explanation

Ans: 10

 To find the \(y\)-coordinate of the \(y\)-intercept of the graph of \(y = 3^x + 9\), we set \(x = 0\) and solve for \(y\):

\[y = 3^0 + 9 = 1 + 9 = 10\]

So, the \(y\)-coordinate of the \(y\)-intercept is \(10\).

Question

\(E = 18,000 – 2,000t\)

        \(V = 18,000(0.85^{t})\)
The given equations are two different models that can be used to find the value, in dollars, of a particular car \(t\) years after it was purchased. Which of the following statements correctly compares the values of \(E\) and \(V\) for \(O < t < 9\) ? 

  1. \(E\) is always less than \(V\).
  2. \(E\) is always greater than \(V\).
  3. \(E\) is initially greater than \(V\) but eventually becomes less than \(V\).
  4. \(E\) is initially less than \(V\) but eventually becomes greater than \(V\).
▶️Answer/Explanation

C

Question

 Kao measured the temperature of a cup of hot chocolate placed in a room with a constant temperature of70 degrees Fahrenheit ( $\left.{ }^{\circ} \mathrm{F}\right)$. The temperature of the hot chocolate was $185^{\circ} \mathrm{F}$ at 6:00 p.m. when it started cooling. The temperature of the hot chocolate was 156 $6^{\circ} \mathrm{F}$ at 6:05 p.m. and $135^{\circ} \mathrm{F}$ at 6:10 p.m. The temperature of the hot chocolate continued to decrease. Of the following functions, which best models the temperature $T(m)$, in degrees Fahrenheit, of Kao’s hot chocolate $m$ minutes after it started cooling? 
A. $T(m)=185(1.25)^m$
B. $T(m)=185(0.85)^m$
C. $T(m)=(185-70)(0.75)^{\frac{m}{5}}$
D. $T(m)=70+115(0.75)^{\frac{m}{5}}$

▶️Answer/Explanation

Ans: D

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