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9 days ago

8

There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in

dance class. Find
When two classes meet at different hours and 12 students are enrolled in both activities. ( 2marks)

When two classes meet at the same hour. ( 2 marks)

1 answer:

zzz [9K]9 days ago

8 0

Answer:

a. 80 students

b. 92 students

Step-by-step explanation:

Designate arts students as A and dance students as D.

Thus, we have,[['[TAG_14]]

n(A) = 35

n(D) = 57

Required

To determine n(A or D)

For (a):

We find that:

n(A and D) = 12

The calculation for n(A or D) is as follows:

n(A or D) = n(A) + n(D) - n(A and D)

n(A or D) = 35 + 57 - 12

n(A or D) = 80

b. Given the details

n(A and D) = 0 since no students are enrolled in both classes as indicated in (a)

Using the same formula as in (a).

n(A or D) = n(A) + n(D) - n(A and D)

n(A or D) = 35 + 57 - 0

n(A or D) = 92

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Answer:

$(A)\dfrac{5s^7}{6}\\(B)s^5\\(E)-6rs^5 \\(F)\dfrac{4r}{5^6}$

Step-by-step explanation:

A polynomial is in its standard form when arranged in descending order based on the variable involved.

Looking at the provided polynomial

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s is presented in descending powers
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$(A)\dfrac{5s^7}{6}\\(B)s^5\\(E)-6rs^5 \\(F)\dfrac{4r}{5^6}$

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The water level of a river is 170 feet. The river recedes 4 feet each year. Ingrid claims that the equation that represents this

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Hank and Debra each own two milking cows. One day, they milked their cows and compared the amount of milk the cows produced in t

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Answer:

The result to your inquiry is 2 7/24.

Step-by-step explanation:

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Debra's cows yield = 5 1/2 and 5 2/3

Process:

1.- Convert the mixed numbers into improper fractions:

For Hank: (16 + 3)/4 = 19/4 and (32 + 1)/8 = 33/8.

For Debra: (10 + 1)/2 = 11/2 and (15 + 2)/3 = 17/3.

2.- Add Hank's and Debra's milk quantities:

Hank: 19/4 + 33/8 = 38/8 + 33/8 = 71/8.

Debra: 11/2 + 17/3 = 33/6 + 34/6 = 67/6.

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67/6 - 71/8 = (268 - 213)/24.

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(78,104) represents the point closest to the interior.

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The equation defining the circle,

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Since the point lies on the circle, its coordinates must be,

$(x,\sqrt{16900-x^2})$

The distance "d" from the point to (30,40) can be calculated as,

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}$

$=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}$

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Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when $9400-60x-80\sqrt{16900-x^2}$ is at its lowest value.

Let’s set up the equation,

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$\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}$

We find the critical points,

$\Rightarrow f'(x)=0$

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$\Rightarrow 80x=60\sqrt{16900-x^2}$

$\Rightarrow 80^2x^2=60^2(16900-x^2)$

$\Rightarrow 6400x^2=3600(16900-x^2)$

$\Rightarrow \dfrac{16}{9}x^2=16900-x^2$

$\Rightarrow \dfrac{25}{9}x^2=16900$

$\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78$

$\Rightarrow x=78$

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$\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2$

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

$\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104$

This leads us to conclude that the closest point is (78,104)

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