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3 years ago

the snake tours are $25 per hour. Write a function that represents the cost S of a tour at Snake Tours for h hours.

Mathematics

1 answer:

Serjik [45]3 years ago

6 0

Answer:

g(h) = 25 + s

(maybe)

Step-by-step explanation:

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Miguel and three of his friends went to the movies. They originally had a total of $40. Each boy had the same amount of money an

Alenkasestr [34]

Answer:

The money each boy had after buying his ticket is $2.50.

Step-by-step explanation:

Given:

Miguel and three of his friends went to the movies. They originally had a total of $40. Each boy had the same amount of money and spent $7.50 on a ticket.

Now, to find the money each boy had after buying his ticket.

Let the money each boy had after buying his ticket be $x.$

Total number of boys = 4.

Total money they had = $40.

Money each boy spent = $7.50.

Now, to write an equation:

$4(x+7.50)=40$

Now, to solve the equation for getting the money each boy had after buying his ticket:

$4(x+7.50)=40$

$4x+30=40$

Subtracting both sides by 30 we get:

$4x=10$

Dividing both sides by 4 we get:

$x=\$2.50.$

Therefore, the money each boy had after buying his ticket is $2.50.

3 0

2 years ago

The minimum/maximum value of the function y = a(x − 2)(x − 1) occurs at x = d, what is the value of d?

daser333 [38]

Answer:

$d=\frac{3}{2}=1.5$

Step-by-step explanation:

We have the function:

$y=a(x-2)(x-1)$

And we want to find x=d for which the minimum/maximum value will occur.

Notice that our function is a quadratic in factored form.

Remember that the minimum/maximum value always occurs at the vertex point.

And remember that the x-coordinate of the vertex is the axis of symmetry.

Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.

Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.

Our factors are (x-2) and (x-1).

Therefore, our roots/zeros are x=1, 2.

So, the average of them are:

$d=\frac{1+2}{2}=3/2=1.5$

Therefore, regardless of the value of a, the minimum/maximum value will occur at x=d=1.5.

Alternative Method:

Of course, we can also expand to confirm our answer. So:

$y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a$

The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.

And the formula for the vertex is:

$x=-\frac{b}{2a}$

So, we will substitute -3a for b and a for a. This yields:

$x=-\frac{-3a}{2a}=\frac{3}{2}=1.5$

Confirming our answer.

4 0

2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$