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Finger [1]
3 years ago
10

Find the distance between points a(2,4) and b(-4,-3 round answer to nearest tenth

Mathematics
1 answer:
vesna_86 [32]3 years ago
3 0
For this, you'd need to know the Euclidean Distance Formula (it's basically the most common one)
                  −−−−−−−−−−−−−−−−
distance=√<span>(<span>x2</span>−<span>x1</span><span>)^2</span>+(<span>y2</span>−<span>y1</span><span>)^2

You would just plug in the points so A is at (2,4) this means that 2 is x1 and 4 is y1. Same with point B but instead of -4 being x1 it would be x2 alsong with -3 being y2.

Ok so when you plug in the points, you get this equation that should be simplified

</span></span>                  −−−−−−−−−−−−−−−−
distance=√(-4−2)^2+(-3−4)^2


                  −−−−−−−−−−−−−−−−
distance=√(-4+(-2))^2+(-3+(-4))^2

I negated the 2 and 4 because of the rules with negatives which basically states that if a positive number is being subtracted from a negative number, it will be multiplied by -1 to make the positive a negative number. Also the subtraction sign will be changed into an addition sign. 
Simplify this. 


                  −−−−−−−−−−−−−−−−
distance=√(-6)^2+(-7)^2


6^2 = 36
7^2= 49

                  −−−−−
distance=√36+49

36+49=85
                  −−
distance=√85

Then the square root of 85 is <span>9.21954445729, but because we are rounding to the nearest tenth</span> the distance is 9.2.


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Pls Answer..........................................
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Answer:

C= 27

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A 10 foot ladder is leaning against a wall. Call x the distance from the top of the ladder to the ground, and call y the distanc
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Step-by-step explanation:

We start by applying the Pythagorean theorem to the ladder, with its length L as the hypotenuse:

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We know that

\dfrac{dy}{dt} = 4\:\text{ft/s}

when x = 6 ft. So the rate at which the top of the ladder is going down is

\dfrac{dx}{dt} = -\left(\dfrac{\sqrt{100\:\text{ft}^2 - (6\:\text{ft})^2}}{6\:\text{ft}}\right)(4\:\text{ft/s})

\:\:\:\:\:\:\:= -5.3\:\text{ft/s}

The negative sign means that the distance x is decreasing as y is increasing.

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