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Nookie1986 [14]
3 years ago
11

What is the slope of the line that passes through the points (3, 2) and (-1,-4)? m=

Mathematics
2 answers:
german3 years ago
5 0

Answer:

3/2

Step-by-step explanation:We know that,

Slope of the straight line passing through the points (x1,y1) and (x2,y2) is

m=y2-y1/x2-x1

Here,

Given, x1= 3,x2= 1

    and y1= 2,y2= -4

so, the slope of the straight line is,

m= -4-2/-1-3

   =3/2

ad-work [718]3 years ago
3 0
The answer is 3/2

Explanation: the formula is y-y /x-x
So it’ll be 2-(-4) /3-(-1) which will give 6 /4 and that simplifies will give 2/ 3
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On a map to cities are about 50 mm apart if the scale is 1 cm equals 15.5 miles how far apart are the two cities
Mademuasel [1]

Given:

On a map to cities are about 50 mm apart.

Scale is 1 cm equals 15.5 miles.

To find:

The distance between two cities.

Step-by-step explanation:

We have,

1 cm = 15.5 miles

It means

10 mm = 15.5 miles             [1 cm = 10 mm]

Now,

10 mm on a map = 15.5 miles

1 mm on a map = \dfrac{15.5}{10} miles

                          = 1.55 miles

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Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

5 0
3 years ago
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