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3 years ago

14

A 1/4 mile long suspension bridge is being planned which will require 16 miles of 150 strand (150 wires twisted ) cable to be pl

aced. What is the minimum (ignoring length for twisting) length in km of wire the cable manufacturer needs to have to produce the cable?

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

out of all of them multimode cables is the best answer

Step-by-step explanation:

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What are numbers that are between 1 and 20 and have exactly 4 factors

vova2212 [387]

Answer:

8 has four factors

Step-by-step explanation:

{1,2,4,8}

1x8=8

2x4=8

Hope this helps!!

3 0

3 years ago

Use pictures and Numbers to show all the ways to take apart 8

Nastasia [14]

2+2+2+2 2×4 8×1 8 4×2

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3 years ago

Noah bought a new car costing $25,350 He took out a 5 year loan at a simple interest rate of 6.5% How much interest did Noah pay

liq [111]

Answer: 8238.75

Work Shown:

i = P*r*t

i = 25350*0.065*5

i = 8238.75

6 0

2 years ago

Read 2 more answers

A wildlife biologist catches, tags, and then releases a number of wasps. Later she catches 55 wasps and finds that 7% are tagged

vesna_86 [32]

Answer:

3.85 tagged wasp

Step-by-step explanation:

6 0

2 years ago

Jenna saves $2,500 per year in an account that earns 10% interest per year, compounded annually. Jenna will have saved in 30 yea

Vsevolod [243]

$~~~~~~~~~~~~\textit{Future Value of an ordinary annuity}\\ ~~~~~~~~~~~~(\textit{payments at the end of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array}\\ pymnt=\textit{periodic payments}\dotfill &2500\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &30 \end{cases}$

$A=2500\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 30}-1}{\frac{0.1}{1}} \right]\implies A=2500\left[ \cfrac{(1.01)^{30}-1}{0.1} \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A\approx 411235.06~\hfill$

4 0

2 years ago