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Lyrx [107]
3 years ago
6

1.

Mathematics
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

Solve it yourself

Step-by-step explanation:

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Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g
SpyIntel [72]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
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 \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see    \bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad  B
\end{array}

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
8 0
3 years ago
Read 2 more answers
Help fast worth 10points
Nina [5.8K]

Answer:

30 ( I think so)

Step-by-step explanation:

Well, the total measurement 4 a triangle is 90. So 60 and 60 is 120 so it should be 30. If not I am sorry for being wrong.

7 0
2 years ago
Kbhhkb,hbj,bhjbj,bj,bh
trasher [3.6K]

Answer:

ahahhajajsiajaklaoaoa

4 0
3 years ago
The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2 : 3, what is the rati
Ksivusya [100]
<h2>Answer:</h2>

The ratio of the area of region R to the area of region S is:

                    \dfrac{24}{25}

<h2>Step-by-step explanation:</h2>

The sides of R are in the ratio : 2:3

Let the length of R be: 2x

and the width of R be: 3x

i.e. The perimeter of R is given by:

Perimeter\ of\ R=2(2x+3x)

( Since, the perimeter of a rectangle with length L and breadth or width B is given by:

Perimeter=2(L+B) )

Hence, we get:

Perimeter\ of\ R=2(5x)

i.e.

Perimeter\ of\ R=10x

Also, let " s " denote the side of the square region.

We know that the perimeter of a square with side " s " is given by:

\text{Perimeter\ of\ square}=4s

Now, it is given that:

The perimeters of square region S and rectangular region R are equal.

i.e.

4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}

Now, we know that the area of a square is given by:

\text{Area\ of\ square}=s^2

and

\text{Area\ of\ Rectangle}=L\times B

Hence, we get:

\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}

and

\text{Area\ of\ Rectangle}=2x\times 3x

i.e.

\text{Area\ of\ Rectangle}=6x^2

Hence,

Ratio of the area of region R to the area of region S is:

=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}

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poizon [28]

Answer:

well- you cant buy any pounds-

Step-by-step explanation:

4 0
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