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3 years ago

The graph of a sinusoidal function intersects its midline at (0,5) and then has a maximum point at (0.75,7).

Mathematics

1 answer:

EastWind [94]3 years ago

5 0

Answer: y(x) = 2*sin(4.19*x) + 5

Step-by-step explanation:

We know that the midline is at x = 0 and y = 5, and the maximum is at x = 0.75 and y = 7

The midline of a graph is a horizontal line that cuts our graph in the middle.

For a normal cosine or sine function, the middle value is zero, so the middle line would be at y = 0, but here we have the midline at (0, 5) so it is located at y = 5.

This means that we have a constant in our function, so it is:

y(x) = f(x) + 5

where f(x) is a trigonometric function.

y(0) = 5 = f(0) + 5

so f(0) = 0

Now we know that sin(0) = 0

Then we have that f(x) = A*Sin(c*x) where A and c are constants.

Now, the maximum of our function is at x = 0.75

and we know that the maximum of the sin(x) is at x = pi = 3.14

then we have:

c*0.75 = 3.14

c = 3.14/0.75 = 4.19

and we have that:

f(0.75) = 7 = A*sin(4.19*0.75) + 5 = A + 5

A + 5 = 7

A = 7 - 5 = 2

Then our function is:

y(x) = 2*sin(4.19*x) + 5

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GREYUIT [131]

Answer:

C, 15

Step-by-step explanation:

i looked it up

5 0

3 years ago

A bus driver pays all tolls, using only nickels and dimes, by throwing one coin at a time into the mechanical toll collector.

lutik1710 [3]

Answer: a) An = An-1 + An-2

b) 55ways

Step-by-step explanation:

a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.

Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that

An = 5n cents

Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;

An-1 = 5( n-1)

Case 2: using a dime which is two 5 cents, the number of ways is given as;

An-2 = 5(n-2)

Summing up the number of ways, we have

An = An-1 + An-2

From the relation,

If n= 0, Ao= 1

n =1, A1= 1

b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)

From the relation, we have that

Ao = 1

A1 = 1

An =An-1 + An-2

Ao = 1

A1 = 1

A2 = A1+Ao = 1+1= 2

A3 = A2 + A1 = 3

A4 = A3+A2=5

A5=A4+A3=8

A6=A5+A4=13

A7 =A6+A5 = 21

A8= A7+A6= 34

A9= A8+A7= 55

So there are 55ways to pay 45cents.

4 0

3 years ago

Y = 3/2x + 3 The slope and y-intercept

ra1l [238]

Answer:

Slope=3/2

Y-intercept=3

Step-by-step explanation:

7 0

2 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

I hope that works for you!

4 0

3 years ago

Pls help soon:)

Tanzania [10]

Correct answers are:
(1) 28, 141 known cases
(2) 79913.71 known cases after six weeks (round off according to the options given)
(3) After approx. 9 weeks (9.0142 in decimal)

Explanations:
(1) Put x = 0 in given equation
y= 28, 141 (1.19)^x
y= 28, 141 (1.19)^(0)
y= 28, 141

(2) Put x = 6 in the given equation:
y= 28, 141 (1.19)^x
y= 28, 141 (1.19)^(6)
y= 79913.71

(3) Since
y= 28, 141 (1.19)^x
And y = 135,000

135,000 = 28, 141 (1.19)^x
135,000/28, 141 = (1.19)^x

taking "ln" on both sides:
ln(4.797) = ln(1.19)^x

ln(4.797) = xln(1.19)
x = 9.0142 (in weeks)

5 0

3 years ago