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2 years ago

Problem Solving

Mathematics

1 answer:

N76 [4]2 years ago

7 0

you're absolutely correct.

each system is sold for $2150, that includes cost + markup, namely the markup is the surplus amount otherwise called "profit".

they sold 12 of those, 2150 * 12 = 25800

they had $4824.36 in profits from it, so if we subtract that from the sale price, we'll be left with the cost of all 12 systems

25800 - 4824.36 = 20975.64

that's the cost for all 12 systems sold, how many times does 12 go into 20975.64? 20975.64 ÷ 12 = 1747.97.

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Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and

SVETLANKA909090 [29]

Answer:k=1,k=-2,k= $8\pm 5\sqrt{3}$

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Given two vectors

$u=1\hat{i}+k\hat{j}$

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$\left ( i\right )$ Distance between them is given by

$|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1$

squaring both side

$1^{2}+\left ( 1-k\right )^2=1$

$k^2-2k+1=0$

$\left ( k-1\right )^2=0$

k=1

$\left ( ii\right )$

angle between u and v is 90 i.e. orthogonal

$u\dot v=0$

$\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )$ =0

2+k=0

k=-2

$\left ( iii\right )$

angle between u & v is $\frac{\pi }{3}$

$u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )$

$|u|=\sqrt{1^2+k^2}$

$|v|=\sqrt{2^2+1^2}$

$2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )$

$\left ( 4+2\right )^2=\left ( 1+k^2\right )5$

$k^2-16k-11$ =0

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3 years ago