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bulgar [2K]
1 year ago
13

A store is currently offering a 60% discount on all items purchased. Your cashier is trying to convince you to open a store cred

it card and says to you, "In addition to the 60% discount you are receiving for purchasing these items on sale today, you will get an additional 20% off for opening a credit card account. That means you are getting 80% off!"
a. What is the mistaken assumption here?
b. Why is that assumption incorrect?
c. If you did truly have 80% discount, explain what should happen when you go to the counter to buy $500 worth of items?
d. If you got your 60% discount and opened the card for an additional 20%, what is the actual % discount you would receive?
e. Is it better to apply the 60% discount first or the 20% discount first?
Mathematics
1 answer:
oee [108]1 year ago
6 0

Answer:

a the assumption is that after the 60% discount, the 20% will contribute to the old price, not the adjusted price

b this is incorrect because the 20% discount will be added on after the 60% discount has been used

c the 500 dollars would turn to 100 dollars

d 40

e it would be best to first apply the 60% coupon first over the 20% coupon

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Find the area of the parallelogram with vertices:________.
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Answer:

97.98

Step-by-step explanation:

The area of the parallelogram PQR is the magnitude of the cross product of any two adjacent sides. Using PQ and PS as the adjacent sides;

Area of the parallelogram = |PQ×PS|

PQ = Q-P and PS = S-P

Given P(0,0,0), Q(4,-5,3), R(4,-7,1), S(8,-12,4)

PQ = (4,-5,3) - (0,0,0)

PQ = (4,-5,3)

Also, PS = S-P

PS = (8,-12,4)-(0,0,0)

PS = (8,-12,4)

Taking the cross product of both vectors i.e PQ×PS

(4,5,-3)×(8,-12,4)

PQ×PS = (20-36)i - (16-(-24))j + (-48-40)k

PQ×PS = -16i - 40j -88k

|PQ×PS| = √(-16)²+(-40)²+(-88)²

|PQ×PS| = √256+1600+7744

|PQ×PS| = √9600

|PQ×PS| ≈ 97.98

Hence the area of the parallelogram is 97.98

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3 years ago
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3 years ago
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Varvara68 [4.7K]

Answer:

The center of circle is (0,0)

Step-by-step explanation:

We need to find the center of the circle of the equation x^{2}+y^{2}=4

Since, the general equation of circle is (x-h)^{2}+(y-k)^{2} = r^{2}

Where (h,k) is center of circle and r is radius.

Re-write the circle equation is x^{2}+y^{2}=4 as,

x^{2}+y^{2}=2^{2}

Compare x^{2}+y^{2}=2^{2} with (x-h)^{2}+(y-k)^{2} = r^{2}

so, (x-0)^{2}+(y-0)^{2} = 2^{2}

Hence, the center of circle is (0,0)

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2 years ago
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