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sergeinik [125]
3 years ago
9

Given the area of a rectangular pool is 6x^3+19x^2+2x-20 and the length of one side of the pool is 2x+5, find the measure of the

missing side.
The missing side length is

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0

Answer:

The missing side length is 3x^2 + 2x -4

Step-by-step explanation:

Here, we are interested in calculating the length of the missing side.

Mathematically, the area of a rectangle = L * B

So since we have the area and one side, to get the other side, we shall need to divide the area by the side we have ;

Thus, the missing side length will be;

6x^3+19x^2+2x-20/2x+ 5

Please check attachment for the polynomial long division

The value of the other factor after the polynomial long division is 3x*2 + 2x -4

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castortr0y [4]
Find a common denominator, then add and simplify

6 0
2 years ago
The product of two positive numbers is 120. One number is 7 more than the other. Determine the two numbers
Sati [7]
Xy= 120
x= 7+y

(7+y)(y) = 120
y^2 + 7y - 120 = 0
(y + 15) (y - 8) = 0

Since we are finding two positive numbers,
If y= 8 then x= 15

The two numbers are 8 and 15
8 0
3 years ago
I need help with #11
Troyanec [42]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

now, with that template in mind, let's take a peek at this function

\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically
8 0
3 years ago
Read 2 more answers
Please answer this correctly.
Ganezh [65]

Answer:

The given function is a nonlinear function because the degree of the function is 2 and we get same value of y for more than one values of x.

Step-by-step explanation:

The given function is

y=2x^2-5

To find the points which lie on the function, put difference values of x in the given function and find the values of y.

Put x= -2

y=2(-2)^2-5=2(4)-5=8-5=3

Put x= -1

y=2(-1)^2-5=2(1)-5=2-5=-3

Put x= 0

y=2(0)^2-5=2(0)-5=-5=-5

Put x=1

y=2(1)^2-5=2(1)-5=2-5=-3

Put x= 2

y=2(2)^2-5=2(4)-5=8-5=3

The table of values is shown below.

Plot these points on a coordinate plane and connect them by a free hand curve.

The given function is a nonlinear function because the degree of the function is 2 and we get same value of y for more than one values of x.

The graph of function is shown below.

5 0
3 years ago
Identify the polygon using as many names as possible
ollegr [7]
A to line b
b to line c
c to line d
d to line e 
e to line a
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