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aniked [119]
3 years ago
12

Given that a function, h, has a domain of -3 ≤ x ≤ 11 and a range of 1 ≤ h(x) ≤ 25 and that h(8) = 19 and h(-2) = 2, select the

statement that could be true for h.
A.

h(13) = 18

B.

h(-3) = -1

C.

h(8) = 21

D.

h(2) = 16

Mathematics
1 answer:
miskamm [114]3 years ago
5 0

Answer:

Actually two points belong to the Domain and Range of this function, namely C, (8,21) D (2,16) and could be true for h(x)

Step-by-step explanation:

1) In other words,  we have two pairs (8,19) and (-2,2):

2) Plugging in each values on every condition:

D=-3 \leqslant x \leqslant 11 \:\:\:R= 1 \leqslant h(x) \leqslant 25\\A)D=-3 \leqslant 13 \leqslant 11 \:\:\:R= 1 \leqslant 18 \leqslant 25\:\:\:\:FALSE\\B)D=-3 \leqslant -3 \leqslant 11 \:\:\:R= 1 \leqslant -1 \leqslant 25 \:FALSE\\C)D=-3 \leqslant 8 \leqslant 11 \:\:\:R= 1 \leqslant 21 \leqslant 25\:True\\D)D=-3 \leqslant 2 \leqslant 11 \:\:\:R= 1 \leqslant 16 \leqslant 25\: True

We have two points that belong to the Domain and Range.

3) Actually two points belong to the Domain and Range of this function, namely C, (8,21) D (2,16) and could be true for h(x)

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Answer:

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Given:}}}}}}}\end{gathered}

  • ⇢ Principle = Rs.4000
  • ⇢ Rate = 6%
  • ⇢ Time = 3 year

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{To Find:}}}}}}}\end{gathered}

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\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Using Formula:}}}}}}}\end{gathered}

{\dag{\underline{\boxed{\sf{Amount  ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}

\dag{\underline{\boxed{\sf{Compound \: Interest = Amount- Principle }}}}

\begin{gathered}\end{gathered}

\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{purple}{Solution:}}}}}}}\end{gathered}

{\bigstar \:{\underline{\pmb{\frak{\red{Firstly,Finding  \: the  \: Amount }}}}}}

\quad {:\implies{\sf{Amount  = \bf{P{\bigg(1  +  \dfrac{R}{100}{\bigg)}^{T}}}}}}

  • Substituting the values

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg(1  +  \dfrac{6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg(1 \times 100  +  \dfrac{6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{100 + 6}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{106}{100}{\bigg)}^{3}}}}}}

\quad {:\implies{\sf{Amount  = \bf{4000{\bigg({\cancel{\dfrac{106}{100}}{\bigg)}}^{3}}}}}}

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\quad {:\implies{\sf{Amount  = \bf{4000{\bigg( \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50}{\bigg)}}}}}}

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\quad {:\implies{\sf{Amount  = \bf{4000 \times  \dfrac{148877}{125000}}}}}

\quad {:\implies{\sf{Amount  = \bf{4{\cancel{000}} \times  \dfrac{148877}{125{\cancel{000}}}}}}}

\quad {:\implies{\sf{Amount  = \bf{\dfrac{148877 \times 4}{125}}}}}

\quad {:\implies{\sf{Amount  = \bf{\dfrac{595508}{125}}}}}

\quad {:\implies{\sf{Amount  = \bf{\cancel{\dfrac{595508}{125}}}}}}

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