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vladimir1956 [14]
3 years ago
13

(9x2 - 8x + 4) - (4x2 + 3x - 9) A. 13x2 - 11x + 5 B. 13x2 - 11x + 13 C. 5x2 - 5x + 13 D. 5x2 - 11x + 13

Mathematics
1 answer:
Troyanec [42]3 years ago
4 0
The answer is d) 5x2-11x+13
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4 0
3 years ago
Meg rowed her boat upstream a distance of 45 mi and then rowed back to the starting point. The total time of the trip was 14 hou
Montano1993 [528]
Recall your d = rt.

so hmm is about the same as the other one... so say the boat has a speed rate of "b", well, going downstream the boat goes say " b + 2 ", because the current is adding to it, and going upstream is " b - 2 " because the current is eroding from it.

So if the whole trip took 14 hours, say it took "t" hours going upstream, then going downstream it took the slack or " 14 - t ".

Also notice, the trip forth and back is the same 45 miles.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Upstream&45&b-2&t\\
Downstream&45&b+2&14-t
\end{array}
\\\\\\
\begin{cases}
45=t(b-2)\implies \frac{45}{b-2}=\boxed{t}\\\\
45=(b+2)(14-t)\\
----------\\
45=(b+2)\left( 14- \boxed{\frac{45}{b-2}}\right)
\end{cases}
\\\\\\
\cfrac{45}{b+2}=14- {\cfrac{45}{b-2}}\implies \cfrac{45}{b+2}=\cfrac{14(b-2)~-~45}{b-2}

\bf \cfrac{45}{b+2}=\cfrac{14b-28-45}{b-2}\implies \cfrac{45}{b+2}=\cfrac{14b-73}{b-2}
\\\\\\
45(b-2)=(b+2)(14b-73)\implies 45b-90=14b^2-45b-146
\\\\\\
0=14b^2-90b-56\implies 0=7b^2-45b-28
\\\\\\
0=(7b+4)(b-7)\implies b=
\begin{cases}
-\frac{4}{7}\\\\
\boxed{7}
\end{cases}
6 0
3 years ago
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