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kondor19780726 [428]
2 years ago
10

If n(U) = 360, n(A) = 240, n(B) = 160, find the maximum value of

Mathematics
1 answer:
Ainat [17]2 years ago
5 0

Answer:

here is answer

Step-by-step explanation:

n(AUB) = n(A) + n(B) - n(AnB )

= 240 + 160 - n(A n B)

= 400 - n(AnB )

Since n(AnB) 20, so n(AUB) S 400

But n(U) = 360, so n(A U B)* 360

.. the maximum value of n(AUB) = 360.

When n(A U B) is maximum, n(AnB) will be minimum.

.. min. value of n(An B)

= n(A) + n(B) - Max, value of n(AUB)

= 240 - 160 - 360 = 40

Again, n(

A B) will be maximum when B C A.

..the max. value of n(

A B) = n(B) = 160

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The above means that, the equation of the cross section passes through the x-axis at:

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y = a(x - 3) * (x + 3)

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Substitute 1/6 for a in y = a(x^2 - 9)

y = 1/6(x^2 - 9)

Hence, the quadratic function that models the cross-section is y = 1/6(x^2 - 9)

Read more about quadratic functions at:

brainly.com/question/1497716

7 0
2 years ago
I got a 12 discount on a pair of jeans and a 25% discount on a t-shirt the original price of the jeans was $100 and thatof the T
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but I'm not sure

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In 1985, the cost of clothing for a certain familiy was $620. In 1995, 10 years later, the cost of clothing for this famility wa
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Answer:

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Increase per year= 380/10

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Second step is to calculate the cost of the clothing in 1991

y = 38*6 + 620

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Therefore the cost of this familiy's clothing in 1991 will be $848

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