A six-meter-long ladder is leaned against a building. If the ladder makes
2 answers:
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Answer:
Option 3: (5, 1)
Step-by-step explanation:
There are several ways of solving for systems of linear equations, either through graphing, substitution, or elimination.
The graphing method is the easiest, as you only need to plot points on the graph using the y-intercepts and the slopes.
I will be demonstrating the <u>graphing method</u>.
Given the systems of linear equations:
Equation 1: y = x - 4
where the <u>slope</u>, <em>m</em> = 1, and <u>y-intercept </u>(0, -4).
Equation2: y = -x + 6
where the <u>slope</u>, <em>m </em>= -1, and <u>y-intercept</u> (0, 6).
The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates of (0, b).
<h3>Graphing Instructions:</h3>For Equation 1, plot the y-intercept, (0, -4) on the graph. Then use the <u>slope</u>, m = 1 (rise 1 unit, run 1 unit) to plot other points on the graph. Repeat this process until you have enough points to connect a line with.
Repeat these same procedures for graphing Equation 2, start by plotting its y-intercept at (0, 6), and its slope, m = -1 (down 1 unit, run 1 unit).
<h3>Solution: </h3>
The intersection of both lines occurs at point, (5, 1), which is solution to the given systems of linear equations. Therefore, the correct answer is the third option, (5, 1).
Attached is the screenshot of the graphed systems of linear equations, where it shows the point of intersection at (5, 1).
I graphed the given points and based on my graph.
<span> x y </span>→ EXPONENTIAL<span><span>
</span> <span> -2 21
</span><span> 1 4
</span><span> 0 1
</span><span> 1 0
</span></span><span> x y </span>→ LINEAR<span><span>
</span> <span> -2 6
</span> <span> -1 -2
</span><span> 0 2
</span><span> 1 6
</span></span>
<span> x y </span>→ QUADRATIC<span><span>
</span> <span> -2 12
</span> <span> -1 -2
</span><span> 0 -8
</span><span> 1 -6
</span></span><span> <span><span> x y </span></span></span>→ EXPONENTIAL<span><span><span>
</span> <span> -2 16
</span> <span> -1 8
</span><span> 0 4
</span><span> 1 2
LINEAR FUNCTIONS always maintain a constant slope. In the above linear function, the slope is 4.
EXPONENTIAL and QUADRATIC FUNCTIONS have different slope in every point.
Exponential functions can be identified by the immediate increase or decrease in the value of y resulting to a minimal increase or decrease in the value of x.
Quadratic functions can be identified as such due to the similar points it has after having the vertex.
</span></span></span>
<span> x y </span>→ EXPONENTIAL<span><span>
</span> <span> -2 21
</span><span> 1 4
</span><span> 0 1
</span><span> 1 0
</span></span><span> x y </span>→ LINEAR<span><span>
</span> <span> -2 6
</span> <span> -1 -2
</span><span> 0 2
</span><span> 1 6
</span></span>
<span> x y </span>→ QUADRATIC<span><span>
</span> <span> -2 12
</span> <span> -1 -2
</span><span> 0 -8
</span><span> 1 -6
</span></span><span> <span><span> x y </span></span></span>→ EXPONENTIAL<span><span><span>
</span> <span> -2 16
</span> <span> -1 8
</span><span> 0 4
</span><span> 1 2
LINEAR FUNCTIONS always maintain a constant slope. In the above linear function, the slope is 4.
EXPONENTIAL and QUADRATIC FUNCTIONS have different slope in every point.
Exponential functions can be identified by the immediate increase or decrease in the value of y resulting to a minimal increase or decrease in the value of x.
Quadratic functions can be identified as such due to the similar points it has after having the vertex.
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