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3 years ago

Pants $80 buy one at regular pric. Get a second one for half price. (How much for 2 pair?)

Mathematics

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

$200 will be price of 2 pairs

Step-by-step explanation:

It is given that the price of one pant is $80

The second is given at half price which is $40

but in the case of pants buying or anything buying when we are given an off then after buying an item the same discount is given for all

so as he bought one pant then the other pant will be given at half price

now the third pant will also be given at half price i.e: $40

fourth price will also be given at $40

so four pants will become 2 pairs

price of 2 pairs = $80+$40+$40+$40

=$200

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ABDF is a regular hexagon. What is the measure of angle A?

dolphi86 [110]

$720 \div 6 = 120$

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2 years ago

Jason went shopping he bought a watch and a pair of trainers for a total price of £53.55 this price includes 15% loyalty discoun

babymother [125]

Answer: price of the watch before the discount is £23.5825

Step-by-step explanation:

Let x represent the original price of the watch.

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15/100 × 53.55 = 8.0325

This means that the original cost of both items is

53.55 + 8.0325 = 61.5825

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3 years ago

Find the midpoint M of the line segment joining the points A= (-7, -6) and B= (1,8).

Marrrta [24]

Step-by-step explanation:

Use midpoint formula,

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3 years ago

Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9

taurus [48]

From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
$\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)$
and
$\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)$

Angle θ is given by
$\theta=180-M-N \\ \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)$

Given that a = 4 units, b = 5 units, and c = 9 units, thus
$\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)$

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
$\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\ \\ -4(x^2+81)+9(x^2-10x+41)=0 \\ \\ -4x^2-324+9x^2-90x+369=0 \\ \\ 5x^2-90x+45=0 \\ \\ x^2-18x+9=0 \\ \\ x=9\pm6 \sqrt{2}$

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
$x=9-6 \sqrt{2}=0.5147$

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.

6 0

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goblinko [34]

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