Martell had boxes he was stacking.Each was 2 feet high.If he stacks 3 boxes on top of a table that is 3 feet high,what will be t
2 answers:
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9514 1404 393
Answer:
see below
Step-by-step explanation:
It is easiest to compare the equations when they are written in the same form.
The first set can be written in slope-intercept form.
y = 2x +7
y = 2x +7 . . . . add 2x
These equations are <em>identical</em>, so have infinitely many solutions.
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The second set can be written in standard form.
y +4x = -5
y +4x = -10
These equations <em>differ only in their constant</em>, so have no solutions.
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The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.
The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11
⇒x=11t
Area of square=
Area of square={as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
Learn more about The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11
⇒x=11t
Area of square=
Area of square={as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
Learn more about area here:
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