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vova2212 [387]
3 years ago
14

The city plans a roadway to have trees every 1/6 mile. If the path is 5 1/2 miles long, how many trees will the city need?

Mathematics
1 answer:
Lana71 [14]3 years ago
4 0

Answer: The city will need 33 trees.

Step-by-step explanation:

Hi, to answer this question we simply have to divide the length of the path (5 1/2 miles) distance between trees (1/6).

Mathematically speaking:

5 1/2 ÷ 1/6 = (5x2+1) /2 ÷1/6 = 11/2 ÷1/6 = (11x6) /(2x1) = 66/2 =33 trees

The city will need 33 trees.

Feel free to ask for more if needed or if you did not understand something.  

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A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet
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Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

h(x)=10+0.1(x^{1.5})

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}

s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx

s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx

s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx

Let 1+0.0225x=t

dx=\frac{1}{0.0225}dt

s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt

s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}

s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}

s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)

s=28.21

Weight of cable=28.21\times 10.4=293.38pound

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3 years ago
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Answer:

By hypotenuse - side test (HL) the two triangles are congruent.

Step-by-step explanation:

In ∆ABC and ∆DCB

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kirill [66]

Answer:

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Step-by-step explanation:

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