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avanturin [10]
3 years ago
5

The possible ways to complete a multiple-choice test consisting of 19 questions, with each question having four possible answers

(a, b, c, or d).
Mathematics
2 answers:
LekaFEV [45]3 years ago
4 0
Let's say there were 3 questions with four answer choices each. That would mean there were 4*4*4 = 16*4 = 64 different ways to answer.

Extend this example out to 19 questions instead of 3. You'll get
4^19 = 4*4*4*...*4*4 = 274,877,906,944
The last value is one big number (not four numbers)
The large number is the answer

note: 4^19 means we have 19 copies of '4' being multiplied together. The three dots mean "continue the pattern". On your paper, you should somehow indicate to your teacher that there are 19 copies of '4' being multiplied.
mr_godi [17]3 years ago
3 0
That would be 4^19 ways which is a huge number.

My calculator give the approximate number as  2.748779069 * 10^11


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The pyramid is shown in the diagram below.
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We have the side of the square, so the area is = 10×10 = 100

We need the height of the triangle to work out its area. We can find out by using the height of the pyramid and half of the length of the side of the square.

Using the Pythagoras rule
Height of triangle = \sqrt{12^{2}+ 5^{2} } =13

Area of one triangle = 1/2×10×12=60
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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1
belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

            F(x) = \int\limits^a_b {f(x)} \, dx

By using integration formula

\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

 \int\limits^6_^-6} (x^{5}  + 8 x^{4}  + 2 x^{2}  + 5 x + 15) )dx

<em>On integration , we get</em>

=   (\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}

F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)

= (\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))

After simplification and cancellation we get

 =  \frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6

on calculation , we get

= \frac{124,416}{5} + \frac{864}{3} + 180

On L.C.M  15

= \frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

6 0
3 years ago
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Andrew [12]

Answer:

the relation between them are

n(U) = n(X U Y) + n(xūy).

4 0
3 years ago
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