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mojhsa [17]
4 years ago
15

 WILL GIVE BRAINLIEST find the distance between points m(5,15) and z(-1,13) to the nearest tenth

Mathematics
1 answer:
arsen [322]4 years ago
3 0
Distance Formula:
\sqrt{  {(x_{2} - x_{1})}^{2}  + {(y_{2} - y_{1})}^{2}}
x_{1} = 5 \\ x_{2} =  - 1 \\ y_{1} = 15 \\ y_{2} = 13
\sqrt{  {(( - 1) - (5))}^{2}  + {((13)- (15))}^{2}} \\  \sqrt{ { - 6}^{2} +  { - 2}^{2}  }  \\  \sqrt{36 + 4}  \\  \sqrt{40}  \\
The Distance between the points is
6.32455532 \\ 6.3 \: to \: nearest \: tenth


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Answer 5 by 6

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A pair of perpendicular lines intersect at the point (5,9). Write
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Answer:

The equation of the line that is perpendicular to the line that passes through the point (-4, 2) is y = -9·x/5 + 18

Step-by-step explanation:

The coordinates of the point of intersection of the two lines = (5, 9)

The coordinates of a point on one of the two lines, line 1 = (-4, 4)

The slope of a line perpendicular to another line with slope, m = -1/m

Therefore, we have;

The slope, m₁, of the line 1 with the known point = (9 - 4)/(5 - (-4)) = 5/9

Therefore, the slope, m₂, of the line 2 perpendicular to the line that passes through the point (-4, 4) = -9/5

The equation of the line 2 is given as follows;

y - 9 = -9/5×(x - 5)

y - 9 = -9·x/5 + 9

y =  -9·x/5 + 9 + 9

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Therefore, the equation of the line that is perpendicular to the line that passes through the point (-4, 2) is y = -9·x/5 + 18.

6 0
3 years ago
I promise I will mark as brainiest if a speed of a car increase its average speed in a journey of 200 miles by 5 mile/hour the j
jok3333 [9.3K]

Answer:

\large \boxed{\sf \ \ \ \dfrac{\sqrt{4025}-5}{2}=29.22144... \ \ \ }

Step-by-step explanation:

Hello

<u>Let's note the original speed of the car v</u>

it means that in 1 hour he is going v miles

so to go 200 miles it takes ( in hour)

\dfrac{200}{v}

If the speed of the car is v+5 than to go 200 miles it takes (in hour)

\dfrac{200}{v+5}

and this time is one hour less so we can write

\boxed{\sf \ \ \dfrac{200}{v+5}=\dfrac{200}{v}-1 \ \ }

We can multiply by v(v+5) both parts of the equation so

200v=200(v+5)-v(v+5)\\\\200v=200v+1000-v^2-5v\\\\v^2+5v-1000=0

\Delta=b^2-4ac=5^2+4*1000=4025\\\\ \text{There are potential solutions }\\\\\ \ \ \ \ x_1=\dfrac{-5-\sqrt{4025}}{2}\\\\\ \ \ \ \ x_2=\dfrac{-5+\sqrt{4025}}{2}

Only one is positive and this is is

x_1=\dfrac{\sqrt{4025}-5}{2}=29.22144...

So the original speed is 29.22144... mph

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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