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Ratling [72]
2 years ago
10

A billboard designer has decided that a sign should have 3 ft margins at the top and bottom and 5 ft margins on the left and rig

ht sides furthermore the billboard should have a total area of 900ft^2 (including the margins) if x denotes the width in feet of the billboard find the function in the variable x giving the area of the printed region of the billboard
Mathematics
2 answers:
elixir [45]2 years ago
7 0

Answer:

The width of billboard is "[x]" and the height of billboard is "[y"]. If total area of billboard is 9000 ft^2 then 9000=xy

Step-by-step explanation:

• The total width of billboard is [x]. Therefore the width of printed area will be (x-10) by excluding margin of left and right side.

• The total height of billboard is [y]. Therefore the height of printed area will be [(y-6)]  by excluding the margin of top and bottom from the total height.

• To find the printed area of billboard calculations are given below:

& 9000=xy

& y=\frac{9000}{x} \\  & A=(x-10)(y-6) \\  & A=xy-6x-10y+60 \\  & A=x\left( \frac{9000}{x} \right)-6x-10\left( \frac{9000}{x} \right)+60 \\  & A=9060-6x-\frac{9000}{x} \\

On taking the first order derivative of A

\[A'=-6+\left( \frac{90000}{{{x}^{2}}} \right)\]

& \left( \frac{90000}{{{x}^{2}}} \right)-6=0 \\          & 6{{x}^{2}}=90000 \\          & x=\sqrt{15000} \\          & y=\frac{9000}{x}=\frac{90000}{\sqrt{15000}}=10\sqrt{150} \\

• Hence \[x=10\sqrt{150}\] and \[y=\frac{900}{\sqrt{150}}\]

Learn More about Differentiation Here:

brainly.com/question/13012860

natta225 [31]2 years ago
6 0

Answer:

aadfddd

Step-by-step explanation:

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nataly862011 [7]

Answer:

The length of the field on the drawing is 55 cm.

Step-by-step explanation:

Given:

Mark made a scale drawing of a soccer field.

Using a scale of .5 cm=1 m.

The actual length of the field is 110 m.

Now, to find the length of the field on drawing.

Let the length of the field on drawing be x.

As given 0.5 cm is equivalent to 1 m.

Thus, x is equivalent to 110 m.

Now, to get the length of the field on drawing by using cross multiplication method:

\frac{0.5}{1} =\frac{x}{110}

<em>By cross multiplying we get:</em>

⇒ 55=x

⇒ x=55

Therefore, the length of the field on the drawing is 55 cm.

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3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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