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?

To find a counterfeit coin among 8 coins with a balance scale, divide the coins and use up to 3 weighings. The algorithm involves grouping and comparing the weights, identifying the counterfeit in the imbalanced group.

Algorithm to Find a Counterfeit Coin among 8 Coins

To find a counterfeit coin among 8 coins with a minimum number of weighings on a balance scale, we need at most 3 weighings. Here is the step-by-step algorithm:

With this algorithm, the maximum number of weighings needed to determine the counterfeit coin is three.

Final answer:

The series 5+25+125+625+3125+15625 can be rewritten in sigma notation as Σ 5*5^(n-1) from n=1 to 6, representing a geometric series with a common ratio of 5 and 6 terms.

Explanation:

To rewrite the series 5+25+125+625+3125+15625 using sigma notation, we first need to identify the pattern of the series. We notice that each term is 5 times the previous term, which is a characteristic of a geometric series. A geometric series can be expressed in sigma notation as Σ a*r^(n-1) from n=1 to N, where a is the first term, r is the common ratio between terms, and N is the number of terms.

In this series, a = 5, and r = 5. The series has 6 terms, so N = 6. Therefore, the series in sigma notation is Σ 5*5^(n-1) from n=1 to 6.

                     Option: A is the correct answer.

                         [tex]x=-2(y+3)^2-1[/tex]

We know that any general equation of a right or left open parabola is given by:

                           x=a(y-k)^2+h

where if a<0 then the parabola open to the left and if a>0 then the parabola opens right.

Also the vertex of the parabola is: (h,k)

Now here we are given:

Vertex (h,k)=(-1,-3)

Also a= -2 in each of the options.

Hence, we have the equation of parabola as:

                [tex]x=-2(y-(-3))^2+(-1)\\\\x=-2(y+3)^2-1[/tex]

Hence, the equation of the parabola is:

                  [tex]x=-2(y+3)^2-1[/tex]

Answer:

d. $10,234.80

Step-by-step explanation:

we are given

The equipment will cost $26,000

so, Amount is $26000

A=26000

should be invested today at 12%

r=12%=0.12

It is compounded semiannually

so, n=2

t=8

now, we can use formula

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]

now, we can plug values

[tex]26000=P(1+\frac{0.12}{2})^{2\times 8}[/tex]

now, we can solve for P

[tex]P=10234.80338[/tex]

Answer:

His rate of discount=10%

Step-by-step explanation:

Step 1: Get the discount amount

Discount amount=List price-sale price

where;

List price=$800

Sale price=$720

replacing;

Discount amount=(800-720)=80

Discount amount=$80

Step 2: Calculate discount rate

The discount rate can be calculated as;

discount rate=(discount/List price)×100

where;

discount=$80

List price=$800

replacing;

discount rate=(80/800)×100=10%

discount rate=10%

We want to answer different things about an exponential decay, the answers are:

So we know that the original price of the computer is $1250 and the value decays by 10% each year.

So in year 1, the new value of the computer will be:

V = $1250*(1 - 0.1) = $1250*0.9

On year 2, the new value will be:

V = ( $1250*0.9)*0.9 = $1250*(0.9)^2

And so on.

a) This sequence is a geometric sequence because each term is a constant times the previous term, where the constant is 0.9.

[tex]A_1 = 1250\\A_2 = 0.9*1250 = 1125\\A_3 = 0.9^2*1250 = 0.9*1125 = 1012.5 \\...[/tex]

b) The explicit formula for a geometric sequence is:

[tex]A_n = k^{n-1}*A_1[/tex]

where:

k is the constant, in this case, is 0.9

A1 is the first term of the sequence, in this case, is 1250

Then we have:

[tex]A_n = (0.9)^{n-1}*1250[/tex]

c) Here we just need to replace n by 6 in the above formula:

[tex]A_6 = (0.9)^5*1250 = 738.1125[/tex]

This means that the price of the computer in the 6th year is $738.10

If you want to learn more about exponential decays, you can read:

https://brainly.com/question/3966275

The value of w is 14.

The image given is a right angled triangle. In order to determine the value of the hypotenuse, w, tan would be used.

Cos 30 = adjacent / hypotenuse

√3 / 2 = 7√3 /  hypotenuse

hypotenuse = 7√3 ÷ √3/2  = 14

To learn more about right angled triangle, please check: https://brainly.com/question/20694080

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Answer:

the answer is p=40h, hope this helps!!!

Step-by-step explanation:

The factorised form of 6m + 18 is given by the equation: 6(m + 3).

Given :

Equation - 6m + 18

Solution :

Factorisation is the method of diminishing the bracket of a quadractic condition, rather than growing the bracket and changing over the condition to a item of variables which cannot be decreased assist.

To factorise the equation 6m + 18 we have to follow some steps:

In first step we have to write 18 = [tex]6\times 3[/tex] so the equation becomes:

[tex]\rm = 6m + (6\times3)[/tex]  ----- (1)

In second step we have to take 6 common from equation (1):

[tex]\rm=6(m+3)[/tex]

So the factorised form of 6m + 18 is given by the equation: 6(m + 3)

For more information, refer the link given below

https://brainly.com/question/19261816

Answer : The equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Explanation :

Half-life = 5.27 years

Formula used :

[tex]a=\frac{a_o}{2^n}[/tex]        ............(1)

where,

a = amount of reactant left after n-half lives

[tex]a_o[/tex] = Initial amount of the reactant

n = number of half lives

And as we know that,

[tex]n=\frac{t}{t_{1/2}}[/tex]        ..........(2)

where,

t = time

[tex]t_{1/2}[/tex] = half-life  = 5.27 years

Now equating the value of 'n' from (2) to (1), we get:

[tex]a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}[/tex]       ...........(3)

[tex]a=\frac{a_o}{2^{(\frac{t}{5.27})}}[/tex]

[tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

Therefore, the equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)^(t/h), where t is the number of years and h is the half-life of the isotope.

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)(t/h), where t is the number of years and h is the half-life of the isotope.

For example, for the radioisotope cobalt-60 with a half-life of 5.27 years, you can use the equation Percent remaining = (100) x (1/2)(t/5.27) to determine the percent remaining after t years.

Let's say you want to find the percent remaining after 10 years, you would substitute t with 10 in the equation, like this: Percent remaining = (100) x (1/2)(10/5.27). Simplify the equation to find the answer.