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REY [17]
3 years ago
13

Luis uses cubes to represent each term of a pattern based on a recursive function. The recursive function defined is f(n + 1) =

f(n) + 4, where n is an integer and n ≥ 2. The number of cubes used in each of the first two figures is shown below. How many cubes does Luis use in the third, fourth, and fifth figures of the pattern? Fill in the blanks.
Figure 1: 9 cubes
Figure 2: 13 cubes
Figure 3:__17____cubes
Figure 4:__21____cubes
Figure 5:_  25____cubes 

NOTICE HOW:  9+4=131+4=17+4=21+4=25 

EACH TIME, EACH NUMBER GOES UP BY 4!!!
Mathematics
2 answers:
damaskus [11]3 years ago
7 0

Answer:

Figure 3: 17 cubes

Figure 4: 21 cubes

Figure 5: 25 cubes

Step-by-step explanation:

Luis uses cubes to represent each term of a pattern based on a recursive function.

The given recursive function defined as

f(n+1)=f(n)+4              .... (1)

where n is an integer and n ≥ 2.

It is given

Figure 1: 9 cubes

Figure 2: 13 cubes

f(1)=9, f(2)=13

Substitute n=2 in equation (1).

f(2+1)=f(2)+4

f(3)=13+4

f(3)=17

Substitute n=3 in equation (1).

f(3+1)=f(3)+4

f(4)=17+4

f(4)=21

Substitute n=4 in equation (1).

f(4+1)=f(4)+4

f(5)=21+4

f(5)=25

Therefore, the number of cubes in figure 3, 4 and 5 are 17, 21 and 25 respectively.

kari74 [83]3 years ago
4 0

Answer:

figure 3: 17 cubes.

figure 4: 21 cubes.

figure 5: 25 cubes.

Step-by-step explanation:

Let the nth figure be represented by f(n).

also we are given f(n+1)=f(n)+4.

The number of cubes used in figure 1 i.e. f(1) is 9.

number of cubes used in figure 2 i.e. f(2) is 13

Now we have to find the number of cube used in figure 3, figure 4 and figure 5.

number of cubes in figure 3 i.e. f(3)=f(2)+4=13+4=17.

number of cubes in figure 4 i.e. f(4)=f(3)+4=17+4=21.

number of cubes in figure 5 i.e. f(5)=f(4)+4=21+4=25.



izzie
3 years ago
I'm sorry but that's wrong
izzie
3 years ago
wait nvm sorry I didn't mean to put it on this one... my bad
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Item 7
Mariulka [41]

Answer:

A = 74.7^\circ

B = 42.5^\circ

C = 62.8^\circ

Step-by-step explanation:

Given

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

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

For AB

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

d = \sqrt{(-1 - 2)^2 + (2 - 8)^2

d = \sqrt{(-3)^2 + (-6)^2

d = \sqrt{45

So:

AB = \sqrt{45

For BC

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

BC = \sqrt{(2 - 4)^2 + (8 - 1)^2

BC = \sqrt{(-2)^2 + (7)^2

BC = \sqrt{53

For AC

A = (-1,2) \to (x_1,y_1)

C = (4,1) \to (x_3,y_3)

AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2

AC = \sqrt{(-5)^2 + (1)^2

AC = \sqrt{26

So, we have:

AB = \sqrt{45

BC = \sqrt{53

AC = \sqrt{26

By representation

AB \to c

BC \to a

AC \to b

So, we have:

a = \sqrt{53

b = \sqrt{26

c = \sqrt{45

By cosine laws, the angles are calculated using:

a^2 = b^2 + c^2 -2bc \cos A

b^2 = a^2 + c^2 -2ac \cos B

c^2 = a^2 + b^2 -2ab\ cos C

a^2 = b^2 + c^2 -2bc \cos A

(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A

53 = 26 +45 - 2 * 34.21 * \cos A

53 = 26 +45 - 68.42 * \cos A

Collect like terms

53 - 26 -45 = - 68.42 * \cos A

-18 = - 68.42 * \cos A

Solve for \cos A

\cos A =\frac{-18}{-68.42}

\cos A =0.2631

Take arc cos of both sides

A =\cos^{-1}(0.2631)

A = 74.7^\circ

b^2 = a^2 + c^2 -2ac \cos B

(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B

26 = 53 +45 -97.67 * \cos B

Collect like terms

26 - 53 -45= -97.67 * \cos B

-72= -97.67 * \cos B

Solve for \cos B

\cos B = \frac{-72}{-97.67}

\cos B = 0.7372

Take arc cos of both sides

B = \cos^{-1}(0.7372)

B = 42.5^\circ

For the third angle, we use:

A + B + C = 180 --- angles in a triangle

Make C the subject

C = 180 - A -B

C = 180 - 74.7 -42.5

C = 62.8^\circ

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