What is the distance between the points (11,19) and (11,-6) in the coordinate plane?

Mathematics

1 answer:

IrinaK [193]1 year ago

6 0

Answer: 25 units

Step-by-step explanation:

distance with negatives is still an addition but you have to turn the negative into a positive to get the answer.

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AlexFokin [52]

The answer is C

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5 0

3 years ago

One more guys. Please make sure you use PEMDAS!

Alexxx [7]

Answer:

-14

Step-by-step explanation:

Replace the 2 for each k.

$\frac{1}{4}2^{5} - 3k^{3} + k^{2} - k$

then solve it using PEDMAS: exponents first

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2 years ago

Resolve into partial fractions<br><br>

lisabon 2012 [21]

Answer:

See below

Step-by-step explanation:

$\frac{11 + x}{(2 - x)(x - 3)} = \frac{A}{2 - x} + \frac{B}{x - 3} \\ \\ \therefore \: \frac{11 + x}{(2 - x)(x - 3)} = \frac{A(x - 3) +B(2 - x) }{(2 - x)(x - 3)} \\ \\ \therefore \: 11 + x = Ax - 3A + 2B - Bx \\ \\ \therefore \: 11 + x = - 3A + 2B + Ax - Bx\\ \\ \therefore \: 11 + x = - 3A + 2B + (A - B)x \\ \\ equating \: the \: like \: terms \: on \: both \: sides \\ A - B = 1 \\ \therefore \: A = B + 1.....(1) \\ \\ - 3A + 2B = 11....(2) \\ from \: eq \: (1) \: and \: (2) \\ - 3(B + 1) + + 2B = 11 \\ \\ - 3B - 3 + 2B = 11 \\ \\ - B = 11 + 3 \\ \\ B = - 14 \\ A = - 14 + 1 = - 13 \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = \frac{ - 13}{2 - x} + \frac{ - 14}{x - 3} \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = - \frac{ 13}{2 - x} - \frac{ 14}{x - 3}$