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3 years ago

HELP help HELP help HELP help HELP help HELP help HELP help HELP help

1 answer:

7 0

$\text{mass of freight train} = 9.98 \times {10}^{6} kg$

$\text{mass of aircraft carrier} = 7.3 \times {10}^{7} \: kg$

$\text{difference of masses of aircraft and train: } \\ \\ 7.3 \times {10}^{7} - 9.98 \times {10}^{6} \\ \\ = 73 \times {10}^{6} - 9.98 \times {10}^{6}$

$= (73 - 9.98 )\times {10}^{6} \\ \\ = 63.02 \times {10}^{6} \\ \\ = 6.302 \times {10}^{6} \: kg \\ \\ \approx 6.3 \times {10}^{6} \: kg$

Approximately, aircraft carrier is 6.3 × 10^6 kg heavier than the freight train.

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Nimfa-mama [501]

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.. h/(3 mi) = tan(θ)
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The rocket's velocity is 18 miles per minute at that moment.

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3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago

A box with an open top will be constructed from a rectangular piece of cardboard. The piece of cardboard is 10 inches wide and 1

kupik [55]

Answer:

Domain of the function is { x : x∈R, x>0 } or (0,∞).

Step-by-step explanation:

It is given that the dimensions of a rectangular piece of cardboard are

Length = 14 inch

Width = 10 inch

The box will be constructed by cutting out equal squares of side x at each corner and then folding up the sides. So the dimensions of the box are

Length = 14-2x inch

Breadth = 10- 2x inch

Height = x inch

The volume of cuboid box is

$V=length\times breadth \times height$

The volume function of the box is

$V=(14-2x)\times (10-2x)\times x$

$V=(14-2x)(10-2x)x$

The volume function is V=(14-2x)(10-2x)x.

It is a polynomial function and domain of a polynomial function is all real numbers.

Here, x represents the height of the box. So, value of x must be a positive real number.

Domain of the function is

Domain = { x : x∈R, x>0 }

It can be written as (0,∞).

Therefore, domain of the function is { x : x∈R, x>0 } or (0,∞).

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