A line has the slope of -1 and a y-intercept of -2. What is an equation that could represent this line?

Mathematics

1 answer:

Shkiper50 [21]2 years ago

3 0

Sorry but I can’t :( I’m 2 dumb ♥️

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Find the solution to the set b^2+12b=0 use a comma

shepuryov [24]

Factor
b(b+12)=0
set to zero
b=0

b+12=0
b=-12

b=0 or -12
written as this way
b=-12,0

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( CORRECT ANSWER GETS BRAINLIEST AND FULL RATINGS !! )

Wewaii [24]

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1) Find the slope of the line passing through the given points (2, 120) and (1, 60).

EastWind [94]

Answer:

1. 60. 2. -6. 3. -4/5

Step-by-step explanation:

Okay so I haven't done slope in 3 years but I used to be a god at it so please tell me if you get this right and I'm so so sorry if you get them wrong

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2 years ago

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GrogVix [38]

The answer would be -2! :)

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3 years ago

F(t)=t^2+19t+60 What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

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2 years ago