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Anton [14]
3 years ago
7

How many times would you have to cross the Golden Gate Bridge in order to travel the exact distance between Wolverhampton and Bi

rmingham?
Mathematics
1 answer:
svetlana [45]3 years ago
4 0

Answer:

Hey!

Well...the distance from Birmingham to Wolverhampton in about 12 miles (19km)

The length of the Golden Gate bridge is 2,737m OR 2.737 km...in miles it's 1.701

So you will have to cross it about 11.2 times (3 sf)

Step-by-step explanation:

19 km / 1.701km = 11.16990005878895

Rounded up to a degree of accuracy = 3 sf

= 11.2

Hope this helps!

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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
Mass of Weight(C)
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I hope this helps
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3 years ago
2. A square-based tent has the cross-sectional
ollegr [7]

(a) Length of the height is 2.732 m

(b) Length of the base is 5.466 m

<u>Explanation:</u>

An image is attached for reference.

(a)

In ΔAOB,

sin 30^o = \frac{AO}{AB} \\\\0.5 = \frac{AO}{2} \\\\AO = 1 m

In ΔBGD,

sin 60^o = \frac{BG}{BD} \\\\0.866 = \frac{BG}{2} \\\\BG = 1.732 m

According to the figure, BG = OE = 1.732 m

Height of the tent, AE = AO + OE

                                  = 1 m + 1.732 m

                                  = 2.732 m

(b)

DF = ?

In ΔAOB,

tan 30^o = \frac{AO}{OB} \\\\0.577 = \frac{1}{OB} \\\\OB = 1.733 m\\\\\\

According to the figure, OB = GE = 1.733 m

In ΔBGD,

tan 60^o = \frac{BG}{DG} \\\\1.732 = \frac{1.732}{DG}\\ \\DG = 1m

According to the figure, DE = DG + GE

                                      DE = 1 m + 1.733 m

                                     DE = 2.733 m

Length of the base, DF = 2 X DE

                              DF = 2 X 2.733 m

                               DF = 5.466 m

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HAL SPENT 5/4 HOUR READING EACH DAY FOR 7 DAYS. HOW MUCH TOTAL DID HE SPEND READING?
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