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3 years ago

7

What is the final product?

Csqrt%7B7%7D%20%20-%203%20%5Csqrt%7B8%7D%20%29" id="TexFormula1" title="(x \sqrt{7} - 3 \sqrt{8} )(x \sqrt{7} - 3 \sqrt{8} )" alt="(x \sqrt{7} - 3 \sqrt{8} )(x \sqrt{7} - 3 \sqrt{8} )" align="absmiddle" class="latex-formula">

1 answer:

Allushta [10]3 years ago

8 0

Hey there !

Check the attachment.
Hope it helps you :)

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PLZ SHOW HOW U WORKED IT OUT

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The leg of a right triangle is 2 units and the hypotenuse is 4 units. What is the length, in units, of the other leg of the tria

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The other leg of the units is 2

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(Algebra II ) What is the equation of y = x^3 with the given transformations? vertical compression by a factor of 1/7, horizonta

NNADVOKAT [17]

$\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}}
\end{array}$ $\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}
\end{array}$

$\bf ----------------------------\\
f(x)=x^3\textit{compression by }\frac{1}{7}\quad or\quad A=7
\\ \quad \\
f(x)=7x^3\impliedby \textit{8 units to the left, or }C=+8
\\ \quad \\
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\\ \quad \\
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3 years ago

M1Q13.) Find the relative frequency of the 2nd class to the nearest tenth of a percent.

erastovalidia [21]

Answer:

9.1%

Step-by-step explanation:

The total number of observations is 33. The number in the second class is 3, so the relative frequency is ...

3/33 × 100% = (100/11)% ≈ 9.1%

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A square flower garden is surrounded by a brick walkway that is 1.8 m wide. The area of the walkway is equal to the area of the

My name is Ann [436]

The length of the garden is 4.3 m while the width is 4.3 m

Let x represent the length of the garden. Therefore:

The area of the garden = x × x = x²

The width of the walkaway is 1.8 m, hence the length of the garden and walkaway = x + 1.8. The width of the garden and walkaway = x + 1.8.

Hence the area of the garden and walkaway = (x + 1.8) × (x + 1.8) = x² + 3.6x + 3.24

The area of walkaway = (x² + 3.6x + 3.24) - x² = 3.6x + 3.24

Since the area of the walkway is equal to the area of the garden, hence:

3.6x + 3.24 = x²

x² -3.6x - 3.24 = 0

x= 4.34 m or -0.74 m

Since the length cannot be negative, hence x = 4.3 m

Therefore the garden is 4.3 m by 4.3 m

Find out more at: brainly.com/question/2077866

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3 years ago