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suter [353]
1 year 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)​
Mathematics
1 answer:
Murrr4er [49]1 year ago
8 0

Answer:

a. 80 students

b. 92 students

Step-by-step explanation:

Represent arts students with A and Dance students with D.

So, we have,

n(A) = 35

n(D) = 57

Required

Determine n(A or D)

Solving (a):

Here, we have:

n(A and D) = 12

n(A or D) is calculated as thus:

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. From the given details

n(A and D) = 0 because both students are not mixed up as in (a) above

Using the same formula as (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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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
1 year ago
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What is the constant of the polynomial 2x3 - 8x2 + 3x - 7
Firlakuza [10]

Answer:

-7

Step-by-step explanation:

A constant number is a number that contains no variables like x and y. The only constant in that problem is -7.

8 0
1 year ago
A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r
bulgar [2K]

Answer:

The answer to your question is 8 h

Step-by-step explanation:

Data

length of the ladder = 200 cm

distance between each rung = 20 cm

rate = 10 cm/h

fifth rung = ?

Process

1.- Calculate the total distance the tide must rise

distance = 20 cm x 4

              = 80 cm              because the first rung touches the water

2.- Calculate the time

rate = distance / time

-Solve for time

time = distance / rate

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time = 80 cm / 10cm/h

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3 0
1 year ago
Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

x_{1} = -5 and x_{2} = 5.

Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

x^{2} > 24

x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

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A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

A = 9.9\,units^{2}

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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