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Savatey [412]
2 years ago
10

Need some help with this question

Mathematics
2 answers:
adelina 88 [10]2 years ago
6 0

Answer:

iba nalang

Step-by-step explanation:

just tell someting wag yan

romanna [79]2 years ago
3 0

1. -

2. ×

3.÷

i guess these r the answers

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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.
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3 years ago
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amm1812

Answer:

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Step-by-step explanation:

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Five pounds is 2.268 times as heavy as one kilogram. How many times as heavy is two pounds than one kilogram?
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7 0
3 years ago
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4
dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

x_1\ =\ -2

y_1\ =\ 3

z_1\ =\ -4

Distance is measure across the line

\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}

So, we can write

\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k

=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k

=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k

=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3

=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18

=>18k-24+12k+9+10k-32\ =\ 18

=>\ k\ =\dfrac{13}{8}

So,

x\ =\ 3k-4

   =\ 3\times \dfrac{13}{8}-4

   =\ \dfrac{7}{4}

y\ =\ \dfrac{4k+3}{2}

   =\ \dfrac{4\times \dfrac{13}{8}+3}{2}

   =\ \dfrac{19}{4}

z\ =\ \dfrac{5k-16}{3}

  =\ \dfrac{5\times \dfrac{13}{8}-16}{3}

   =\ \dfrac{-21}{8}

Now, the distance of point from the plane is given by,

d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}

   =\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}

   =\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}

   =\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}

   =\ \sqrt{\dfrac{1177}{64}}

   =\ 4.28

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

            = 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0
3 years ago
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