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aleksklad [387]
3 years ago
14

(Write your answers on the space provided) The World Health Organization (WHO) reported that about 16 million adolescent girls b

etween 15 and 19 years of age give birth each year. Knowing the adverse effects of adolescent childbearing on the health of the mothers as well as their infants, a group of students from Magiting High School volunteered to help the government in its campaign for the prevention of early pregnancy by giving lectures to 7 Barangays about the WHO guidelines on teenage pregnancy. The group started in Barangay 1 and 4 girls attended the lecture. Girls from other barangays heard about it, so 8 girls attended from Barangay 2, 16 from Barangay 3, and so on. a. Make a table representing the number of adolescent girls who attended the lecture from Barangay 1 to Barangay 7 assuming that the number of attendees doubles each Barangay 4 3 5 6 2 7 1 Barangay Number of Attendees b. Form a sequence representing the number of adolescent girls who attended the lecture from Barangay 1 to Barangay 7. C. Because people who heard about the lecture given by the group thought that it would be beneficial to them, five more different barangays requested the group to do the lectures for them. If the number of young girls who will listen to the lecture from these five Barangays will increase in the same manner as that of the first 7 Barangays, determine the total number of girls who will benefit from the lecture.​

Mathematics
1 answer:
navik [9.2K]3 years ago
5 0

The sequence of the number of girl attendees at each Barangay is nonzero, given by multiplying the number of attendees in the previous Barangay by 2

The required values are;

a. The completed table is presented as follows;

\underline{\begin{array}{|l|c|c|c|c|c|c|c|}\mathbf{Barangay}&1&2&3&4&5&6&7\\\mathbf{Number \ of \ attandees}&4&8&16&32&64&128&256\end{array}}

b. The sequence of the number of girls attendees aₙ = <u>4×2⁽ⁿ⁻¹⁾</u>

c. The sum of beneficiaries of the lecture from the twelve Barangay is <u>16,384 girls</u>

Reason:

Known parameters are;

Number of girls that attended the lecture at Barangay 1 = 4 girls

Number of girls that attended from Barangay 2 = 8 girls

Number that attended from Barangay 3 = 16 girls

a. The required table with the assumption that the number of girls that attended at each Barangay from Barangay 1 to Barangay 7 doubles, is presented as follows;

  • \begin{array}{|l|c|c|c|c|c|c|c|}Barangay&1&2&3&4&5&6&7\\Number \ of \ attandees&4&8&16&32&64&128&256\end{array}

b. The sequence of the number of girls who attended the lecture from Barangay 1 to Barangay 7 is presented as follows;

The sequence is a geometric progression having the form, aₙ = a·r⁽ⁿ⁻¹⁾

The first term of the sequence, a = 4

The common ratio, r = 2 (each term is twice the previous term)

n = The number of terms

The sequence is therefore;

  • <u>aₙ = 4·2⁽ⁿ ⁻ ¹⁾</u>

c. The total number of girls in the geometric progression, (G. P.) is given by the sum of a G. P. as follows

S_n = \dfrac{a \cdot (r^n - 1 )}{r - 1}

Where;

n = The number of terms = 7 + 5 = 12

Which gives;

  • S_{12} = \dfrac{4 \times (2^{12} - 1 )}{2 - 1} = \dfrac{4 \times 4,096 }{1} = 16,384

The total number of girls who will benefit from the lectures, S₁₂ = <u>16,384 girls</u>

Learn more about geometric progression here:

brainly.com/question/14256037

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A quadrilateral is simply a polygon with four sides, four angles, and four vertices.

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Option C) is the correct answer.

Learn more about area of rectangle here: brainly.com/question/27612962

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\gray{ \frak{The \:  given \:  two \:   \: points  \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}

\:

Let's solve by using midpoint formula :

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\bf \boxed{\color{red}\frak{Midpoint \:   \: Formula :  {( \: x ,\:y \:  ) = }(  \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}

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\: \large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}

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\large \gray{ \frak{( \: x ,\:y \:  ) = }(  \frak{  \cancel\frac{ -4}{2} ,  \cancel\frac{ - 5}{2} )}}

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\underline{ \boxed{ \large \red{ \frak{option \: d }(  \frak{   - 2, - 2.5)}}}}✓

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