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2 years ago

11

A model airplane has a length of 15in and a width of 9in. If that actual airplane is going to be 65 feet long, how wide is the a

ctual plane?
A. 42ft
B. 58ft
C. 39ft
D. 23ft

1 answer:

Vaselesa [24]2 years ago

3 0

The answer would be C.

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You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

8 0

3 years ago

Please help, this is for 8th grade math.

expeople1 [14]

Answer:

(2,-2)

Step-by-step explanation:

The first step is to substitute y in the first equation with the right side of the second equation because they both equal to y.

-3x+4 = 4x-10

-3x-4x+4-4 = 4x-4x-10-4

-7x = -14

-7x/-7 = -14/-7

x = 2

Plugin 2 for x for either equation to solve for y, I'll be using the second equation, but either one is fine.

y = 4(2)-10

y = 8-10

y = -2

Plugin x = 2 and y = -2 into (x,y), and you get (2,-2)

6 0

3 years ago

Write 6 ratinol numbers between - 3 and - 4<br><br>

WARRIOR [948]

Answer:

-3.1, -3.2, -3.3, -3.4, -3.5, -3.6

Step-by-step explanation:

7 0

2 years ago

Lisa buys 3 pints of milk how many cups did Lisa buy?

AnnZ [28]

Answer:

3 pints = 6 cups

Step-by-step explanation:

Formula: multiply the volume value by 2

3 0

2 years ago

Read 2 more answers

Write each of the following as a function of theta.<br> 1.) sin(pi/4 - theta) 2.) tan(theta+30°)

Paladinen [302]

Step-by-step explanation:

Let x represent theta.

$\sin( \frac{\pi}{4} - x )$

Using the angle addition trig formula,

$\sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y)$

$\sin( \frac{\pi}{4} ) \cos(x) - \cos( \frac{\pi}{4} ) \sin(x)$

$( \frac{ \sqrt{2} }{2}) \cos(x) - (\frac{ \sqrt{2} }{2} )\sin(x)$

Multiply one side at a time

Replace theta with x , the answer is

$\frac{ \sqrt{2} \cos(x) }{2} - \frac{ \sin(x) \sqrt{2} }{2}$

2. Convert 30 degrees into radian

$\frac{30}{1} \times \frac{\pi}{180} = \frac{\pi}{6}$

Using tangent formula,

$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) }$

$\frac{ \tan(x) + \tan( \frac{\pi}{6} ) }{1 - \tan(x) \tan( \frac{\pi}{6} ) }$

Tan if pi/6 is sqr root of 3/3

$\frac{ \tan(x) + ( \frac{ \sqrt{3} }{3} ) }{1 - \tan(x) (\frac{ \sqrt{3} }{3} ) }$

Since my phone about to die if you later simplify that,

you'll get

$\frac{(3 \tan(x) + \sqrt{3} )(3 + \sqrt{3} \tan(x) }{3(3 - \tan {}^{2} (x) }$

Replace theta with X.

4 0

2 years ago