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3 years ago

Find the Distance between the points (-6,5)and(8, -3)

Mathematics

1 answer:

marusya05 [52]3 years ago

7 0

Answer:

$d=16.1245$

Step-by-step explanation:

Distance Formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Simply plug in your 2 coordinates into the formula to find distance d:

$d=\sqrt{(8-(-6))^2+(-3-5)^2}$

$d=\sqrt{(8+6)^2+(-8)^2}$

$d=\sqrt{(14)^2+64}$

$d=\sqrt{196+64}$

$d=\sqrt{260}$

$d=16.1245$

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Answer Question number 5 Plzzzzzz

loris [4]

Answer:

Step-by-step explanation:

a = -4/3

d= -1 +4/3 =1/3

last term = 4 1/3 = 13/3

nth term = a +(n-1)d

$\frac{-4}{3}+(n-1)*\frac{1}{3}=\frac{13}{3}\\\\(n-1)*\frac{1}{3}=\frac{13}{3}-\frac{-4}{3}\\\\(n-1)*\frac{1}{3}=\frac{13+4}{3}\\\\(n-1)*\frac{1}{3}=\frac{17}{3}\\\\n-1=\frac{17}{3}*\frac{3}{1}\\\\n-1=17\\\\n=17+1=18$

Middle terms are 9the term and 10th term

$t_{9}=\frac{-4}{3}+8*\frac{1}{3}=\frac{-4}{3}+\frac{8}{3}=\frac{4}{3}\\\\t_{10}=\frac{-4}{3}+9*\frac{1}{3}=\frac{-4}{3}+\frac{9}{3}=\frac{5}{3}\\\\ t_{9}+t_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3\\\\t_{9}+t_{10}=3$

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