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3 years ago

Find the area of a square that has a perimeter of 24 cm.

Mathematics

2 answers:

aniked [119]3 years ago

6 0

Answer: 36cm²

Step-by-step explanation:

24÷4=6

6×6=36

Colt1911 [192]3 years ago

5 0

Answer:36

Step-by-step explanation: since a square has 4 sides you divide 24 by 4 to see what one side equals which is 6 then you do 6 times 6 because that's how you find area of a square which gets you 36

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3 years ago

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What is an example of a square root function? In (,) form

RUDIKE [14]

Step-by-step explanation:

The parent function of the functions of the form f(x)=√x−a+b is f(x)=√x . Note that the domain of f(x)=√x is x≥0 and the range is y≥0 . The graph of f(x)=√x−a+b can be obtained by translating the graph of f(x)=√x to a units to the right and then b units up.

I hope it's helpful!

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3 years ago

Find the scale factor please.

torisob [31]

Answer:

5/2

Step-by-step explanation:

The scale factor for linear measures, such as perimeter, is the square root of the scale factor for areas. The ratio of larger area to smaller is ...

... 75/12 = 25/4

so the ratio of the larger perimeter to the smaller is ...

... √(25/4) = 5/2

3 0

3 years ago

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

3 years ago