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3 years ago

What is the slope of a line that passes through the point (4,5) and (-6,-3)

Mathematics

1 answer:

Mkey [24]3 years ago

5 0

M= y1-y2/x1-x2

m= 8/10

m= 4/5

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BRAINLIEST! BRAINLIEST! PERMUTATION Find the value: (show your solution)

dlinn [17]

Answer:

$(1) \ _8P_8 = 40320\\\\(2) \ _1_0P_4 = 5040\\\\(3) \ \frac{_1_0P_5}{_5P_3} = 504\\\\(4) \ \frac{_6P_2 \ . \ _8P_5}{_1_0P_6} = 1\frac{1}{3} \\\\(5) \ \frac{_8P_2 \ . \ _8P_6}{_7P_6 \ . \ _4P_2} = 18\frac{2}{3}$

Step-by-step explanation:

$(1) \ _8P_8 = \frac{8!}{(8-8)!} = \frac{8!}{0!} = 8! = 40320$

$(2) \ _1_0P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} = \frac{10\times 9\times 8 \times 7 \times 6!}{6!} = 10\times 9\times 8 \times 7 = 5040$

$(3) \ \frac{_1_0P_5}{_5P_3} =\frac{(10!)/(10-5)!}{(5!)/(5-3)!} = \frac{(10!)/(5!)}{(5!)/(2!)} = \frac{10\times 9 \times 8\times 7\times 6}{5\times 4 \times 3} = 504$

$(4) \ \frac{_6P_2 \ . \ _8P_5}{_1_0P_6} = \frac{(6!)/(6-2)! \ \times\ (8!)/(8-5)!}{(10!)/(10-6)!} = \frac{(6!)/(4!) \ \times \ (8!)/(3!)}{(10!) /(4!)} = \frac{6! \ \times \ 8!}{10! \ \times \ 3!} \\\\= \frac{6\times 5 \times 4 \times 3! \ \times 8!}{10\times 9 \times 8! \ \times \ 3!} = \frac{6\times 5 \times 4}{10\times 9} = \frac{4}{3} = 1\frac{1}{3}$

$(5) \ \frac{_8P_2 \ . \ _8P_6}{_7P_6 \ . \ _4P_2} = \frac{(8!)/(6!) \ \times \ (8!)/(2!)}{(7!)/(1!) \ \times \ (4!)/(2!)} = \frac{8! \ \times \ 8!}{7! \ \times \ 6! \ \times \ 4!} = \frac{(8 )\ \times \ (8 \times 7)}{4!} =\frac{(8 )\ \times \ (8 \times 7)}{4\times 3 \times 2}\\\\= \frac{56}{3} = 18 \frac{2}{3}$

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PLEASE HELP!!! Write the equation of the line that is perpendicular to y = -3/2x + 1 and passes through (0,-4), please show me s

Y_Kistochka [10]

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3 years ago

What is the slope of a line that passes through the point (4,5) and (-6,-3)​

What is the slope of a line that passes through the point (4,5) and (-6,-3)