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2 years ago

Write 12/15 in the simplest form

2 answers:

5 0

If you would like to know what is 12/15 in the simplest form, you can calculate this using the following step:

12/15 simplifies to 4/5 (the common factor of 12 and 15 is 3, so you can divide both numbers by 3).

The result is 4/5.

snow_lady [41]2 years ago

4 0

I believe that is is 4/5

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Sixteen is the same as one and two-tenths times a number added to negative 8

Archy [21]

Answer:

Step-by-step explanation:

16=1 2/10R + -8

16+8= 12/10R

24=12/10R

R=20

3 0

3 years ago

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Write answer as x= x-21=99

murzikaleks [220]

X=120
This is the answer because 120-21=99

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2 years ago

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Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D′E′F′G′H′I′:

dalvyx [7]

Answer:

Step-by-step explanation:

it goes down by 9 and left by 3

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2 years ago

What is the distance between the points (5.-18) and (8,

romanna [79]

Answer:

$\sqrt{10}$

Step-by-step explanation:

The horizontal distance from points (5,-18) and (8,-17) is 3 because it is 3 units from 5 to 8. The vertical distance is 1 since it is one unit from -18 to -17. Now we can use the equation $a^{2}+b^{2}=c^{2}$ where a=3 and b=1 and c is the distance that you are looking for:

$a^{2} +b^{2} =c^{2}\\3^{2} +1^{2} =c^{2}\\9+1=c^{2}\\10=c^{2}\\\sqrt{10}=c$

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2 years ago

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write a function g whose graph represents a translation 2 units to the right followed by a horizontal stretch by a factor or 2 o

Alenkinab [10]

$\bf f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}\\\\
f(x)=&{{ A}} \left|{{ B }}x+{{ C}} \right|+{{ D}}
\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

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

with that template in mind, let's see,

$\bf f(x)=|x| \implies \begin{array}{lllccll}
f(x)=&1|&1x&+0|&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.

the smaller A is, the wider it opens, the larger it is, the more it shrinks.

5 0

2 years ago