I’d love to help I pretty sure it is C-6
why is no one answering? I also need help in this question!!!!
Yes, your answer is correct
Answer:
(-4, 3)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations by substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
-3x - 3y = 3
y = -5x - 17
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: -3x - 3(-5x - 17) = 3
- Distribute -3: -3x + 15x + 51 = 3
- Combine like terms: 12x + 51 = 3
- Isolate <em>x</em> term: 12x = -48
- Isolate <em>x</em>: x = -4
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = -5x - 17
- Substitute in <em>x</em>: y = -5(-4) - 17
- Multiply: y = 20 - 17
- Subtract: y = 3
Answer:
- <em>The actual temperature after the freezer was on for five minutes could be </em><u><em>18ºF or 12ºF</em></u>
Explanation:
First, you must find the y-value returned by the line in the graph for a time (x) of five minutes, and then use the deviation of 3º to conclude the range of the real temperature.
<u>1. y- value at x = 5 min.</u>
The trend line (blue line in the graph) passes through the point (5,15), which means that it predicts a temperature of 15ºF when the time (x) is 5 minutes, or five minutes after the freezer was turned on.
<u>2. Actual temperature value</u>
It is stated that the temperature was actually three degrees from what the trend line shows, that means that the actual temperature could be 3ºF more or 3ºF less than the predicted temperature of 15ºF.
Mathematically:
- Temparature = 15ºF + 3ºF or 15ºF - 3ºF
- Temperature = 18ºF or 12ºF ← answer.