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olga nikolaevna [1]
3 years ago
8

Each part of a race is 1/10 mile long. Marsha finished 5 part of the race.How far did marsha race?

Mathematics
1 answer:
andre [41]3 years ago
5 0

We can multiply the number of parts Marsha has finished by the miles of each part:

5 × 1/10

= 5/10

=1/2

Therefore, she raced 1/2 miles.

Hope it helps!

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Answer:

breadth = 8, length = 32

Step-by-step explanation:

perimeter = 2 x (length x breadth)

                = 2 x ((24 + b) + b))

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                = 48 + 4 b

48 + 4b meters costs rupees 2400

each meter costs rupees 30

so,

30 x (48 + 4b) = 2400

1440 + 120b    = 2400

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Find how many quarts of 6% butterfat milk and 1% butterfat milk should be mixed to yield 75 quarts of 3% butterfat milk.
IgorC [24]

Answer:

30 gallons of 6% milk, and 45 gallons of 1% milk.

Step-by-step explanation:

x=quarts of 6% butterfat

<h2>First, write an equation.</h2>

0.06x+0.01(75-x)=0.03*75

<h2>Multiply both sides by 100.</h2>

0.06x*100+0.01*100(75-x)=100*0.03*75

<h2>Simplify</h2>

6x+75-x=3*75

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<h2>Solve</h2>

5x=225-75=150

5x=150

x=30

75-30=45

So, we need 30 gallons of 6% milk, and 45 gallons of 1% milk.

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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)

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\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}

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3 years ago
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