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Len [333]
3 years ago
7

Car A and car B are traveling at different speeds. Car A can travel 175 miles in the same time it takes car B to travel 125 mile

s. If car A travels 20 mph faster than car B, how fast (in mph) is car B traveling? (Enter an exact number.)
Mathematics
1 answer:
slega [8]3 years ago
8 0

Answer:

The speed of car B is 50 mile per hour

Step-by-step explanation:

Given as :

The Distance travel by car A = 175 miles

The Distance travel by car B = 125 miles

The Time taken by both car A and B is same = T hours

Let The speed of car B = S mile per hour

So, The speed of car A = ( S + 20 ) mile per hour

Now, Time = \frac{Distance}{Speed}

So , For Car A ,

T = \frac{175}{(S + 20 )}

For Car B ,

T = \frac{125}{(S}

So, \frac{175}{(S + 20 )}  = \frac{125}{(S}

Or, 175 × S = 125 × ( S + 20 )

Or, 175 S - 125 S = 125 × 20

Or, 50 S = 2500

∴  S = \frac{2500}{(50}

I.e S = 50 mph

And S + 20 = 50 + 20 = 70 mph

The speed of car A = 70 mph

The speed of car B = 50 mph

Hence The speed of car B is 50 mile per hour  Answer

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Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

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Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

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\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

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\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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