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3 years ago

A shopkeeper marks the price of an article

Mathematics

1 answer:

marin [14]3 years ago

3 0

Answer:

The gain percent made by shopkeeper is 10.5%.

Step-by-step explanation:

Given:

Price at which goods are marked = 30%

Discount on the marked price = 15%

To Find:

Gain percentage of the shopkeeper

Solution:

Let the cost price of the product be = 100

Therefore,

MP = 30 + 100 = 130

Since discount is 15%. thus -

Discount = 15/100 x 130

= 19.5

Selling price = MP - Discount

SP = 130 - 19.5

= 110.5

Profit = SP - CP

= 110.5 - 100

= 10.5

Calculating the gain%

Gain% = 10.5/100 x 100

= 10.5

Answer: The gain percent made by shopkeeper is 10.5%.

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Geometry problem help! Please refer to the image below...

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Answer:

A. 1/3

B. √10

C. -1, 1

D. √8, 6

E. congruent and opposite pairs parallel

F. perpendicular, not congruent

G. rhombus, explanation below

Step-by-step explanation:

Hey there! I'm happy to help!

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Slope is the rise over the run. Let's look at F to G.

We are going from -1 to 2 on our x-axis (run), so our run is 3 units.

Our rise is 1 unit as we go from 2 to 3 on the y-axis.

$slope=\frac{rise}{run} =\frac{1}{3}$

This slope is the same for all of the sides.

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We will use the distance formula (which is basically just the Pythagorean Theorem) to calculate the length of each side. Let's go between F and G again, but this distance is the same for all the sides.

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-1,2)\\\\(x_2,y_2)=(2,3)\\\\\\\sqrt{(2+1)^2+(3-2)^2 } \\\\\sqrt{(3)^2+(1)^2 }\\\\\sqrt{9+1 }\\\\\sqrt{10}$

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The diagonals are the lines that connect the non-adjacent vertices.

Our two diagonals are FH and GE.

-----------------------------

FH

We go from x-value -1 to 1 from F to H, so our run is 2.

We go from y-value 2 to 0. so our rise is -2.

$slope=\frac{rise}{run} =-\frac{2}{2} =-1$

-----------------------------

GE

We go from x-value -2 to 2 from E to G, so our run is 4.

We go from y-value -1 to 3. so our rise is 4.

$slope=\frac{rise}{run} =\frac{4}{4} =1$

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Let's use the distance formula on each of our diagonals.

-----------------------------

FH

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-1,2)\\\\(x_2,y_2)=(1,0)\\\\\\\sqrt{(1+1)^2+(0-2)^2 } \\\\\sqrt{(2)^2+(-2)^2 }\\\\\sqrt{4+4 }\\\\\sqrt{8}$

-----------------------------

GE

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-2,-1)\\\\(x_2,y_2)=(2,3)\\\\\\\sqrt{(2+2)^2+(3+1)^2 } \\\\\sqrt{(4)^2+(4)^2 }\\\\\sqrt{16+16 }\\\\\sqrt{36}\\\\6$

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They are congruent as they all have the same length (√10) and the opposite sides are parallel as they have the same slope (1/3)

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They are perpendicular diagonals as their slopes are negative reciprocals (1 and -1), and they are not congruent as they have different lengths (√8 and 6).

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Parallelogram- quadrilateral with opposite pairs of parallel sides.

Rhombus- a parallelogram with four equal sides

Square- a rhombus with four right angles

We can see that this is a parallelogram as we saw that the opposite sides are parallel due to having the same slope, and the perpendicular diagonals show that as well. This is also a rhombus because if we use that distance formula on all the sides, it will be the same. It is not a square though because it does not have four right angles, so this is a rhombus.

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Have a wonderful day and keep on learning!

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