Answer:
Step-by-step explanation:
Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:
Using the distance formula, we get
Similarly,
Now, the length of the rectangle is and the width of the rectangle is . So, the area of the rectangle is:
Therefore, the area of the rectangle is 20 square units.
Step-by-step explanation:
55.44=2464
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Answer:
y=3/4x-3
Step-by-step explanation:
Okay so to be parallel to an equation of another line, the line must have the same slope but a different y-intercept. First, you need to find the slope of the equation -3x+4y=4. The equation must be in y=mx+b format, so you first move -3x to the right by adding it on both sides, leaving you with 4y=3x+4. Next, you divide 4y by 4 in order to isolate y and do that on the other side. Now, you have y=3/4x+1.
To find the equation of the new line, you must put the point and slope into point slope form: y-y1=m(x-x1). In the point (4, 0), 4 would be x1 and 0 would be y1. so, the new equation is y-0=3/4 (x-4). now, distrubute the 3/4, leaving you with y=3/4x-3. This is correct since the slope remains the same in both equations with a different y-intercept. To check deeper, you could place point (4, 0) on a graph and then rise 3, run 4 until you get the line :) hope this helps
Answer:
a. f(-1)=12
b. f(2t)=16t²-6t+5
c. f(t-2)=4t²-19t+27
Step-by-step explanation:
For a, b, c, we are given an input. We plug that into f(x) to find our answers.
a. f(-1)=12
f(-1)=4(-1)²-3(-1)+5
f(-1)=4+3+5
f(-1)=12
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b. f(2t)=16t²-6t+5
f(2t)=4(2t)²-3(2t)+5
f(2t)=4(4t²)-6t+5
f(2t)=16t²-6t+5
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c. f(t-2)=4t²-19t+27
f(t-2)=4(t-2)²-3(t-2)+5
f(t-2)=4(t²-4t+4)-3t+6+5
f(t-2)=4t²-16t+16-3t+6+5
f(t-2)=4t²-19t+27
