Question

1. graph the line described by the equation -2x - 4y = 16 (images of four graphs labeled a, b, c, d)
2. an object is thrown upward with an initial velocity of 35 meters per second. the object’s distance, d, above the ground at any time, t, can be represented by the equation d = 35t - 5t². when will the object be 50 feet above the ground?

a. t = 1 sec and t = 0.4 sec c. t = 2 sec and t = 10 sec
b. t = 2 sec and t = 5 sec d. t = 5 sec and t = 10 sec

1. jasmine and her sister are saving to buy mp3 players. jasmine has $50 and plans to save $10 per week. her sister has $80 and plans to save $7 per week. in how many weeks will jasmine have more money saved than her sister?

a. 2 weeks c. 10 weeks
b. 4 weeks d. 11 weeks

Explanation:

Problem 51

Step1: Rewrite the equation in slope - intercept form

The given equation is \(-2x - 4y=16\). We want to solve for \(y\) in terms of \(x\).
First, add \(2x\) to both sides of the equation: \(-4y = 2x + 16\).
Then, divide each term by \(-4\): \(y=-\frac{2}{4}x-\frac{16}{4}\), which simplifies to \(y =-\frac{1}{2}x - 4\).
The slope of the line is \(-\frac{1}{2}\) and the \(y\) - intercept is \(- 4\).

Step2: Analyze the \(y\) - intercept and slope

The \(y\) - intercept is \(-4\), so the line crosses the \(y\) - axis at \((0,-4)\). The slope is \(-\frac{1}{2}\), which means for every 2 units we move to the right along the \(x\) - axis, we move down 1 unit.
Now, let's check the graphs:

• For graph a: Check the \(y\) - intercept. If the \(y\) - intercept is not \(-4\), we can eliminate it.
• For graph b: The \(y\) - intercept is \(-4\) (since it crosses the \(y\) - axis at \((0, - 4)\)) and the slope is \(-\frac{1}{2}\) (the line is decreasing with a slope of \(-\frac{1}{2}\)).
• For graph c: Check the \(y\) - intercept. If it is not \(-4\), eliminate.
• For graph d: Check the \(y\) - intercept. If it is not \(-4\), eliminate.

Step1: Set up the equation

We know that \(d = 35t-5t^{2}\) and we want to find \(t\) when \(d = 50\). So we set up the equation \(35t-5t^{2}=50\).

Step2: Rearrange the equation to standard quadratic form

Subtract 50 from both sides: \(-5t^{2}+35t - 50 = 0\). Multiply both sides by \(- 1\) to make the coefficient of \(t^{2}\) positive: \(5t^{2}-35t + 50=0\).

Step3: Simplify the quadratic equation

Divide each term by 5: \(t^{2}-7t + 10 = 0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to 10 and add up to - 7. The numbers are - 2 and - 5. So, \(t^{2}-7t + 10=(t - 2)(t - 5)=0\).

Step5: Solve for \(t\)

Set each factor equal to zero:

• \(t - 2=0\) gives \(t = 2\) seconds.
• \(t - 5=0\) gives \(t = 5\) seconds.

Step1: Set up the inequality

Let \(w\) be the number of weeks. Jasmine's savings after \(w\) weeks is \(S_{J}=50 + 10w\) (she starts with $50 and saves $10 per week). Her sister's savings after \(w\) weeks is \(S_{s}=80+7w\) (she starts with $80 and saves $7 per week). We want to find \(w\) when \(S_{J}>S_{s}\).
So, \(50 + 10w>80 + 7w\).

Step2: Solve the inequality

Subtract \(7w\) from both sides: \(50+3w>80\).
Subtract 50 from both sides: \(3w>30\).
Divide both sides by 3: \(w > 10\). Wait, this seems wrong. Wait, maybe the problem is "Jasmine has $50 and plans to save $10 per week. Her sister has $80 and plans to save $7 per week. In how many weeks will Jasmine have more money saved than her sister?" Wait, maybe I misread the problem. Wait, maybe it's "Jasmine has $50 and plans to save $10 per week. Her sister has $80 and plans to save $7 per week. In how many weeks will Jasmine have as much or more?" Wait, let's re - do the inequality:
\(50 + 10w>80+7w\)
\(10w-7w>80 - 50\)
\(3w>30\)
\(w > 10\). But the options are 2,4,10,11. Wait, maybe the problem is "Jasmine has $50 and plans to save $10 per week. Her sister has $80 and plans to save $7 per week. In how many weeks will Jasmine have more money saved than her sister?" Wait, if \(w = 11\), \(S_{J}=50+10\times11 = 160\), \(S_{s}=80 + 7\times11=80 + 77 = 157\). \(160>157\). For \(w = 10\), \(S_{J}=50 + 10\times10=150\), \(S_{s}=80+7\times10 = 150\). So at \(w = 11\), Jasmine has more.

Answer:

Graph b (assuming the labels are as per the standard, the graph with \(y\) - intercept \(-4\) and slope \(-\frac{1}{2}\))

Problem 52