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Naya [18.7K]
2 years ago
14

Which is the equation of the line with slope 3 that contains point (-1,5)?

Mathematics
2 answers:
Elena-2011 [213]2 years ago
7 0
The answer is y-5=3(x+1)
kotegsom [21]2 years ago
4 0

Answer:

D

y-5=3(x+1)

Step-by-step explanation:

We can see that the question is asking for the Point-Slope Form;

y-y_{1} =m(x-x_{1})

The points go into the equation for y_{1} and x_{1}

y-5=m[x-(-1)]

M would be the slope;

y-5=3[x-(-1)]

Which basically is:

y-5=3(x+1)

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the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function
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The function of the area of the square is A(t)=121t^{2}

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

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Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

Learn more about The function of the area of the square is A(t)=121t^{2}

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

\frac{dx}{dt} = 11 \frac{cm}{sec}

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Area of square=(length of side)^{2}

Area of square=(11t)^{2}{as the length of side is 11t}{varies by time}

Area of square=121t^{2}

Therefore,The function of the area of the square is A(t)=121t^{2}

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