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nata0808 [166]
3 years ago
11

The number of adults who attend a county fair (measured in hundreds of people) is represented by the function a(d)=−0.3d2+4d+9 ,

where d is the number of days since the fair opened. The number of children who attend the same county fair (measured in hundreds of people) is represented by the function c(d)=−0.2d2+5d+11 , where d is the number of days since the fair opened. What function, f(d) , can be used to determine how many more children than adults attend the fair on any day?
Mathematics
2 answers:
denis-greek [22]3 years ago
7 0
<span> f(d)=0.1d2+d+2 would be your answer I believe</span>
Lesechka [4]3 years ago
6 0

Given is the function for number of adults who visit fair at day 'd' after its opening, a(d) = −0.3d² + 4d + 9.

Given is the function for number of children who visit fair at day 'd' after its opening, c(d) = −0.2d² + 5d + 11.  

Any function f(d) to find excess of children more than adults can be written as follows :-  

f(d) = c(d) - a(d).

⇒ f(d) = (−0.2d² + 5d + 11) - (−0.3d² + 4d + 9)

⇒ f(d) = -0.2d² + 0.3d² + 5d - 4d + 11 - 9

⇒ f(d) = 0.1d² + d + 2

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(¬‿¬) Find the number of real number solutions for the equation. x^2+9x+20=0
mart [117]

To find the nature of solutions of a quadratic equation we can find the discriminant.

The standard form of a quadratic equation is ax² + bx + c = 0.

Formula to find the discriminant = b² - 4ac.

So, first step is to compare the given equation with the above equation to get the value of a, b, and c.

After comparing the two equations we will get a =1, b = 9 and c = 20.

Next step is to plug in these value in the formula of discriminant. So,

Discriminant = 9² - 4(1)(20)

= 81 - 80

= 1

Since 1 is a positive perfect square number.

So, we will get two real solutions for this equation.

Hence, the correct choice is C.

Hope this helps you.

8 0
3 years ago
Please help, its so simple
antoniya [11.8K]

Answer:

117

Step-by-step explanation:

Subtract from 180 to get interior and exterior angles.

180 - 101 = 79

79 + 38 = 117

180 - 117 = 63

180 - 63 = 117

5 0
3 years ago
Read 2 more answers
Solve for x:a(a²+b²)x²+b²x-a​
m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a  

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

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

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b²  = (b² + 2a²)²

2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

5 0
3 years ago
How to graph {3x-8y = 29
labwork [276]
<span>How to graph {3x-8y = 29 3x+y = -2}
</span> The first thing you should do is to clear "y". Then, you must substitute values of x in the equation to find the values of "y". Finally, graph the points obtained. (see graphic)
 How do you convert this to slope intercept form? 3x-8y = 29 ---> y = 0.375x - 3.625 3x+y = -2 ---> y = -3x - 2

5 0
3 years ago
Question1) Describe three scenarios that involve a real-world linear or exponential function. At least one must be exponential.
elixir [45]

Answer:

1) Let's suppose that you go in a straight line, in a car that moves at a constant speed of 80km/h.

Then the distance from your house (assuming that you start the drive in your house) can be modeled with a linear equation:

D(t) = 80km/h*t

where t is time in hours.

This will be a linear function.

2) Suppose that you have a population of some animal, that grows by 2% each month, and initially, there are 100 individuals of that animal.

Then the first month, the population is 100.

The second month the population increased by a 2%, then it will be:

100 + 100*0.02 = 100*(1.02)

The third month, the population will be 100*(1.02) + 0.2*100*(1.02) = 100*(1.02)^2.

and so on, this is an exponential relation, where the population as a function of the number of months, can be written as:

P(m) = 100*(1.02)^(m - 1)

3) Suppose that you have $100 saved, and each month you can save another $80, let's find a function that says the amount of money that you have saved as a function of the number of months. S(m)

The month number zero (before you started saving) you had $100 saved.

S(0) = $100.

One month after, you have saved $80 more, then you have:

S(1) = $100 + $80

Another month after, you have:

S(2) = $100 + $80 + $80 = $100 + 2*$80

And so on, you already can see the pattern, after m months, you will have:

S(m) = $100 + m*$80 saved.

5 0
3 years ago
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