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Fiesta28 [93]
2 years ago
5

Where are white blue and red boat in the Marina three fourths of the boats in the Marina are white for seventh of the remaining

boats are blueAnd the rest are red if there are nine boats how many boats are in the Marina?
Mathematics
1 answer:
xeze [42]2 years ago
5 0

Answer:

28

Step-by-step explanation:

Here is the complete question used in answering this question :

Three-fourths of the boats in the marina are white. 4/7 of the remaining boats are blue , and the rest are red . If there are 9red boats, how many boats are in the marina?

We first have to determine the fraction of the boats that are red

1 - (\frac{3}{4} + \frac{4}{7})

1 - \frac{21 + 16}{28}

1 - \frac{37}{28}

1 - 1\frac{9}{28}

= 9/28

so, 9/28 of the boats are red

let b represent the total number of boats

9/28 x b = 9

to find b, divide both sides of the equation by 28/9

b = 28

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Eva8 [605]

Answer:  i) 1 - 9x² - 12x

               ii) 17 - 3x²

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<u>Step-by-step explanation:</u>

g(x) = 3x + 2       h(x) = 5 - x²

i) h(g(x))

  h(3x + 2) = 5 - (3x + 2)²

                  = 5 - (9x² + 12x + 4)

                  = 5 - 9x² - 12x - 4

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ii) g(h(x))

   g(5 - x²) = 3(5 - x²) + 2

                 = 15 - 3x² + 2

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iii) h(h(x))

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                 = 5 - (25 - 10x² + x⁴)

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4 0
3 years ago
A rectangular box which is open at the top can be made from an12-by-18-inch piece of metal by cutting a square from each corner
algol [13]

Answer:

h=2.35in, l=13.3in and w=7.3in

Step-by-step explanation:

In order to solve this problem, we must start by drawing a diagram that will represent the prolem. (Look at attached picture)

From the diagram we can see that the length ad the width of the box are found by  using the following expressions:

L=18-2x

W= 12-2x

H=x

next, we can make use of the volume formula of a rectangular prism, which is the following:

V=WLH

so we can substitute the expressions we got before on the formula:

V=x(12-2x)(18-2x)

and for ease of calculation, we can multiply the terms, so we get:

V=x(216-24x-36x+4x^{2})

V=x(4x^{2}-60x+216)

V=4x^{3}-60x^{2}+216x

So now we can find the derivative of the volume:

V'=12x^{2}-120x+216

and set it equal to zero, so we get:

12x^{2}-120x+216=0

and now we can solve. We can start by factoring the left side of the equation, so we get:

12(x^{2}-10x+18)=0

which yields:

x^{2}-10x+18=0

This one can only be solved by using the quadratic formula:

x=\frac{-b\pm \sqrt{-b^{2}-4ac}}{2a}

so we substitute the corresponding values:

x=\frac{-(-10)\pm \sqrt{-(-10)^{2}-4(1)(18)}}{2(1)}

which yields:

x=7.65in or x=2.35in

we keep the 2.35in only, because the 7.65in returns an impossible box, since that would make the width a negative width.

So the 2.35in height would give us the greatest box. Now we can use it to find the height, the width and the length of the box.

Height=x=2.35

W=12-2x=12-2*2.35=7.3in

L=18-2x=18-2(2.35)=13.3in

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