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3 years ago

How do you convert fraction to a decimal?

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

$\huge\text{Hey there!}$

$\large\text{Some people usually convert fractions into decimals by DIVIDING}\\\large\text{the TOP number) by the DENOMINATOR (the BOTTOM number)}.$

$\large\text{Here is an EXAMPLE of what I meant!}$

$\large\text{Fraction: }\mathsf{\dfrac{5}{6}}$

$\large\text{Equation: } \mathsf{5\div6}$

$\large\text{Answer: } \mathsf{0.833333}$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

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Name one of the two Spelman College professors interviewed in Beyond Beats and Rhymes.

Papessa [141]

Answer:

One of the two professor mentioned in Beyond Beats and Rhymes is William Jelani Cobb

Step-by-step explanation:

Beyond Beats and Rhymes is a documentary movie made in 2006 that was focus on the issues of sexism, homophobia, violence and masculinity in HIP-HOP music genre the movie starred Carmen Ashurst-Watso , Clipse , Fat Joe and other Hip Hop artistes and was directed by Byron Hurt

7 0

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)

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

4 0

3 years ago

The arch beneath a bridge is semi-elliptical, a one-way roadway passes under the arch. The width of the roadway is 38 feet and

forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}$

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

$a=\frac{38}{2}=19ft$

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

$\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}$

and we can simplify the equation, so we get:

$\frac{x^2}{361}+\frac{y^2}{144}$

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

$\frac{y^{2}}{144}=1-\frac{x^{2}}{361}$

$y^{2}=144(1-\frac{x^{2}}{361})$

we can add everything inside parenthesis so we get:

$y^{2}=144(\frac{361-x^{2}}{361})$

and take the square root on both sides, so we get:

$y=\sqrt{144(\frac{361-x^{2}}{361})}$

and we can simplify this so we get:

$y=\frac{12}{19}\sqrt{361-x^{2}}$

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

$y=\frac{12}{19}\sqrt{361-(8)^{2}}$

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.