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VMariaS [17]
3 years ago
7

What is the y-intercept of the equation of the line that is perpendicular to the line y = 3/5 x + 10 and passes through the poin

t (15, –5)?
The equation of the line in slope-intercept form is y = -5/3 x +
Mathematics
1 answer:
TiliK225 [7]3 years ago
8 0

Answer:

\large\boxed{y-intercept=20}

Step-by-step explanation:

\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=\dfrac{3}{5}x+10\to m_1=\dfrac{3}{5}.\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{3}{5}}=-\dfrac{5}{3}.\\\\\text{The equation of the searched line:}\ y=-\dfrac{5}{3}x+b.\\\\\text{The line passes through }(15,\ -5).

\text{Put thecoordinates of the point to the equation.}\ x=15,\ y=-5:\\\\-5=-\dfrac{5}{3}(15)+b\\\\-5=(-5)(5)+b\\\\-5=-25+b\qquad\text{add 25 to both sides}\\\\b=20\\\\\boxed{y=-\dfrac{5}{3}x+20}

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If mountain mack splits his week evenly between carving fishing lures and duck decoys, what is the maximum number of fishing lur
Aleksandr [31]
Given the question:

Mountain Mark and Big Lake Bob spend their time carving fishing lures and duck decoys. The table below shows their production possibility schedule, which describes various combinations of fishing lures and duck decoys they each can carve efficiently in a week.

\begin{center}
\begin{tabular}
{|c {2.3cm}|c {2.3cm}|c {2.3cm}|c {2.3cm}|}
\multicolumn {4} {|c|} {Production Possibility Schedules}\\[1ex]
\multicolumn {2} {|c|} {Mountain Mark}&\multicolumn {2} {|c|} {Big Lake Bob}\\[1ex]
Fishing Lures&Duck Decoys&Fishing Lures&Duck Decoys\\[1ex]
40&0&100&0\\
32&10&80&6\\
24&20&60&12\\
16&30&40&18\\
8&40&20&24\\
0&50&0&30
\end{tabular}
\end{center}

<span>If mountain mack splits his week evenly between carving fishing lures and duck decoys, what is the maximum number of fishing lures and duck decoys he could carve?


From the table it can be seen that in a full week, if mountain mack is carving fishing lures only, he can carve 40 fishing lures. This means that in half of a week, he can carve 20 fishing lures.

Similarly, f</span><span>rom the table it can be seen that in a full week, if mountain mack is carving duck decoys only, he can carve 50 duck decoys. This means that in half of a week, he can carve 25 duck decoys.
</span>
Therefore, i<span>f mountain mack splits his week evenly between carving fishing lures and duck decoys, the maximum number of fishing lures and duck decoys he could carve are 20 fishing lures and 25 duck decoys.</span>
5 0
3 years ago
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