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3 years ago

Make q the subject of the formula 2(q+p) = 1+5q. Give your answer in the form ap-b/c where a,b and c are all positive integers

Mathematics

1 answer:

musickatia [10]3 years ago

8 0

Answer:

$\frac{2p-1}{3}$

Step-by-step explanation:

Since you want to get q only and q appears in both side of the equation. Try to isolate q to one side.

1) Expand 2(q+p)

2q + 2p = 1 + 5q

2) Move all q terms to one side

5q - 2q = 2p - 1

3q = 2p - 1

3) Divide 3 on both side (to isolate q)

q = $\frac{2p-1}{3}$

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Farmer Smith enclosed his rectangular pasture with fence that cost him $1.50 per linear meter. If his pasture is twice as long a

maria [59]

Answer:

$2,700

Step-by-step explanation:

Let the width of the pasture be x, this means that the length of the pasture will be 2x

The area of a rectangle is L * B

Hence,

x * 2x = 18,000

2x^2 = 180,000

x^2 = 90000

x = 300 meters

This means the length is 600 metres

Now, to get the length of the fence, we need to know the actual perimeter which is 2(L + B)

= 2 ( 300 + 600)

2 * 900 = 1,800 metres

The cost is thus 1,800 * $1.50 = $2,700

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3 years ago

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Answer:

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3 years ago

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$