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3 years ago

Given AB= BC FINDIND X= AB= BC= AC=

Mathematics

1 answer:

tester [92]3 years ago

8 0

Answer:

see explanation

Step-by-step explanation:

Given AB ≅ BC, then

AB = BC = x + 6 , and

AC = AB + BC ← substitute values

3x - 31 = x + 6 + x + 6, that is

3x - 31 = 2x + 12 ( subtract 2x from both sides )

x - 31 = 12 ( add 31 to both sides )

x = 43

Hence

AB = BC = x + 6 = 43 + 6 = 49

AC = 3x - 31 =(3 × 43) - 31 = 129 - 31 = 98

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2 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

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

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

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2 years ago

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