What is the length of side BC of the triangle? Make sure to show all of your work!

Mathematics

1 answer:

riadik2000 [5.3K]4 years ago

4 0

BC = 48

<c and <b are congruent (as shown in picture)
AC and AB are congruent because of Base Angles Theorem

Make an equation for the two side lengths:
x + 8 = 2x - 1
x - 1 = 8
x = 9

Then you can plug in for BC
3(17) - 3
51 - 3
48 Units

You might be interested in

the graph of f(x), shown below, has the same shape as the graph of g(x)=x^2, but it is shifted up 3 units and to the right 1 uni

UkoKoshka [18]

A. Takes it 1 to the right and 3 up

8 0

4 years ago

Read 2 more answers

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

$\bf ----------------------------\\
f(x)=x^3\textit{compression by }\frac{1}{7}\quad or\quad A=7
\\ \quad \\
f(x)=7x^3\impliedby \textit{8 units to the left, or }C=+8
\\ \quad \\
f(x)=7(x+8)^3\impliedby \textit{now, reflection over x-axis, means}
\\ \quad \\
f(x)=-7(x+8)^3\impliedby \textit{a negative leading term's coefficient}$

6 0

3 years ago

What is the answer for 6r^2-10r-24

FrozenT [24]

$6r^2-10r-24 =0\\ \\a=6, \ b= -10, \ c= -24 \\ \\ \Delta =b^2-4ac = (-10)^2 -4\cdot6\cdot (-24) = 100+576 = 676 \\ \\ r_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{10-\sqrt{676}}{2\cdot 6 }=\frac{ 10-26}{12}= \frac{-16}{12}=- \frac{4}{3} \\ \\r_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{10+\sqrt{676}}{2\cdot 6 }=\frac{ 10+26}{12}= \frac{36}{12}=3\\ \\ 6r^2-10r-24 = 6(r+\frac{4}{3})(r-3)=(6r+8)(r-3)\\ \\ \\ Answer : \ 6r^2-10r-24 =(6r-8)(r-3)$