Write a polynomial in standard form that represents the area of the shaded region.

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

Answer:

p^2 - 2p - 3.

Step-by-step explanation:

The area of a triangle = 1/2 * base * height so here we have

Area = 1/2 (p + 1)(2p - 6)

= 1/2( 2p^2 - 6p + 2p - 6)

= 1/2 ( 2p^2 - 4p - 6)

= p^2 - 2p - 3.

Kamila [148]3 years ago

4 0

1/2•(2p-6)(p+1)
1/2•(2p^2+2p-6p-6)
1/2•(2p^2-4p-6)
p^2-2p-3

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3 years ago

The following data set represents the age of

Julli [10]

$\huge\text{Hey there!}$

$\huge\textsf{Breaking the meaning down in simpler terms}$
$\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}$

$\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}$

$\huge\textsf{Formulas for each term}$

$\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }$

$\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}$

$\huge\textbf{SOLVING FOR THE QUESTION}$

$\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}$

$\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}$

$\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}$