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3 years ago

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brian buys bikes to resell. he applies a mark up of 15%. write two expressions to represent the selling price. let (b) represent

the original cost of the bike.

1 answer:

AleksandrR [38]3 years ago

6 0

You can do it and if you can't then I will tell you

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Plz tell me the aswer and how you worked it out. A tree grows to a height of 13.75m over 5 1/2 years. What is the average annual

marin [14]

1. Take your 13.75 and 5 1/2 and put it into a dividing table
_______
5.5 I13.75

you need to get your decimal out of deviser (5.5) will now be (55) because you moved the decimal over there you also need to move the decimal that is on the inside to and it will be (137.5)

2. When you divide you would get 2.5

3. Now you would know how much the tree grew over the 5 1/2 years.

ANSWER: 2.5m a year!

Hope this helped!

:)

4 0

3 years ago

A maps key shows that every 5 inches on the map represents 200 miles of actual distance. Suppose the distance between two cities

ryzh [129]

Hello there, and thank you for posting your question here on brainly.

Short answer: 280 miles.

Why?

You can find this out by finding out how much 1 in is by dividing 200 by 5. (200 / 5 ==> 40) So 1 in = 40 miles. Now, we multiply that by 7, so we can find out how much 7 in would be. (7 * 4 ==> 280) 7 inches on the map represents 280 miles.

Hope this helped you! ♥

3 0

3 years ago

I have no idea how to solve this. please help?

matrenka [14]

I believe it’s c

π=C/2r

8 0

3 years ago

A stock brokerage has four computers that are used for making trades on the New York Stock Exchange. The probability that a comp

frutty [35]

Answer:

625/100,000,000 or 6.25 out of one million

Step-by-step explanation:

(0.005)^4 = .000000000625 = 625/100,000,000

5 0

3 years ago

Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g

Anika [276]

Complete Question:

Find the directional derivative of g(x,y) = $x^2y^5$ at the point (1, 3) in the direction toward the point (3, 1)

Answer:

Directional derivative at point (1,3), $D_ug(1,3) = \frac{162}{\sqrt{8} }$

Step-by-step explanation:

Get $g'_x$ and $g'_y$ at the point (1, 3)

g(x,y) = $x^2y^5$

$g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486$

$g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405$

Let P = (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

$u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\$

$|\bar{PQ}| = \sqrt{8}$

The unit vector is therefore:

$u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }$

The directional derivative of g is given by the equation:

$D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }$

8 0

3 years ago