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3 years ago

I need help with this problem

Mathematics

1 answer:

cupoosta [38]3 years ago

7 0

$\bf \begin{cases}
(n^4)^p=n^{12}\to &n^{4\cdot p}=n^{12}\to 4p=12
\\
&\qquad \uparrow \\
&\textit{same bases, thus}\\
&\qquad \downarrow 
\\
n^3\cdot n^q=n^6\to &n^{3+q}=n^6\to 3+q=6
\end{cases}
\\\\\\
thus\implies 
\begin{cases}
4p=12\implies p=\frac{12}{4}
\\\\
3+q=6\implies q=6-3
\end{cases}
\\\\

\\\\
then\implies p\cdot q=\boxed{?}$

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Please help FAST I need the answer along with you explaining the steps so I can get it thanks! I’ll give Brainlist also!

kupik [55]

Answer:

The equation of required line is: $\mathbf{5x-8y+34=0}$

Step-by-step explanation:

We need to write equation in the form Ax+By+C=0 of the line parallel to $5x-8y+12=0$ and through the point (-2,3)

First we need to find slope and y-intercept of the required line.

Using equation of line $5x-8y+12=0$ to find slope.

Since the given line and required lines are parallel there slope is same.

Writing equation in slope intercept form: $y=mx+b$ where m is slope.

$5x-8y+12=0 \\-8y=-5x-12\\\frac{-8y}{-8y}=\frac{-5x}{-8}-\frac{12}{-8}\\y=\frac{5}{8}x+\frac{3}{4}$

So, the slope m = 5/8

The slope of required line is $m=\frac{5}{8}$

Now finding y-intercept using slope $m=\frac{5}{8}$ and point(-2,3)

$y=mx+b\\3=\frac{5}{8}(-2)+b\\3=\frac{-5}{4}+b\\b=3+\frac{5}{4}\\b=\frac{3*4+5}{4}\\b=\frac{12+5}{4}\\b=\frac{17}{4}$

So, the equation of required line having slope $m=\frac{5}{8}$ and y-intercept $b=\frac{17}{4}$

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So, The equation of required line is: $\mathbf{5x-8y+34=0}$

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