Start with 3 one-by-one squares. This represents the '3' in 4t+3
Then draw 4 rectangles that are vertical or horizontal. Make sure the rectangle is longer than it is wide, or vice versa. The longer side is t units long (t is just a placeholder for a number). The shorter side is 1 unit long
The longer thin rectangles have an area of 1*t = t square units. Four of them represent t+t+t+t = 4*t = 4t
The small squares have an area of 1*1 = 1. Three of them represent 1+1+1 = 1*3 = 3
This is one possible way to draw it out. See the attached drawing.
Answer:
Step-by-step explanation:
the last one
Answer: income
Step-by-step explanation:
Answer:
d = -1/3, 0
Step-by-step explanation:
Subtract the constant on the left, take the square root, and solve from there.
(6d +1)^2 + 12 = 13 . . . . given
(6d +1)^2 = 1 . . . . . . . . . .subtract 12
6d +1 = ±√1 . . . . . . . . . . take the square root
6d = -1 ±1 . . . . . . . . . . . .subtract 1
d = (-1 ±1)/6 . . . . . . . . . . divide by 6
d = -1/3, 0
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Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:
f(x) = (6x+1)^ +12 -13
We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.
1. no because 2x+10 is not equal to 2x+7
2. no. you cant add una like terms so 4x+4 doesn’t equal 8x
3. yes. the terms still retain their same value although they are not in the same order
4. yes. when u distribute the -3, you get -3x-6 which is of course equal to -3x-6