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IgorC [24]
2 years ago
7

CHALLENGE PROBLEM Danny's Department Store advertises savings of

Mathematics
1 answer:
Oksanka [162]2 years ago
8 0

Discount are used to reduce the costs of items

The percent paid to determine the advertisement is incorrect

<h3>How to determine the correct percentage</h3>

The initial prices are given as: $44.00 to $152.00

And the discount is: 45% to 51%

At a discount of 45%, the price of the items would be:

Price = \$44.00 * (1 - 45\%)

Price = \$24.2

Price_2 = \$152.00 * (1 - 45\%)

Price_2 = \$83.6

At a discount of 51%, the price of the items would be:

Price = \$44.00 * (1 - 51\%)

Price = \$21.56

Price_2 = \$152.00 * (1 - 51\%)

Price_2 = \$74.48

The calculates prices show that the percent paid to determine the advertisement is incorrect

Read more about discounts at:

brainly.com/question/23103804

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Step-by-step explanation:

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  • Which then gives you y = - 2

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

7 0
3 years ago
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