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3 years ago

Write the equation of the line passing through (-1,0) and (0,-3)

Mathematics

2 answers:

Nuetrik [128]3 years ago

5 0

Answer:

y = -3x - 3.

Step-by-step explanation:

The slope of the line = (y2-y1)/(x2-x1) where (x1, y1) and (x2, y2) are 2 points on the line.

Here x1 = -1 , x2 = 0, y1 = 0 and y2 = -3.

So the slope = (-3 - 0) / (0 - (-1))

= -3/1

= -3.

The equation of a line can be written as

y = mx + c where m = the slope and c = the y-intercept.

So substituting m = -3 and the point (-1, 0):

0 = -3(-1) + c

c + 3 = 0

c = -3.

So our equation is y = -3x - 3.

enot [183]3 years ago

4 0

Answer:

y = -3x -3

Step-by-step explanation:

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Which two equations are true?(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)=6.6×10−10 (6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.

tiny-mole [99]

let me edit your question as:

Which two equations are true?

Eq1:

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

Eq2:

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

Eq3:

2.7×103(7.5×106)−(2.5×106)=5×100

Answer:

No one is true

Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

Eq1:

$(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\$

Eq2:

$6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1$

Eq3:

$2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\$

we observed that none of the equation has two same values on both sides thus none of the three equations is true.

Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true

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