In one U.S. city, the taxi cost is $2 plus $0.90 per mile. If you are traveling from the airport, there is an additional charge of $3.00 for tolls. How far can you travel from the airport by taxi for $18.50

The Greates Common Factor (GCF) is:   2

The value that is equivalent to 6 times 5 squared is 150.

In mathematics, an expression is a combination of numbers, variables, and mathematical operations (such as addition, subtraction, multiplication, division, and exponentiation) that represents a mathematical relationship or a quantity.

Expressions in mathematics can be evaluated or simplified to produce a single value or to represent a more simplified form of the original expression.

6 times 5 squared can be written in a few different ways:

6 x 5²

6 x 25

150

So the expressions that are equivalent to 6 times 5 squared are:

6 x 5²

6 x 25

The value that is equivalent to 6 times 5 squared is 150.

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The answer to  [tex]72 - (-58) - 62[/tex] is 68.

Understanding Double Negatives: When we subtract a negative number, it is equivalent to adding that number. Therefore, [tex]-(-58)[/tex] becomes [tex]+58[/tex].

So, the expression can be rewritten as:

[tex]72 + 58 - 62[/tex]

Performing the Addition: Now we add [tex]72[/tex] and [tex]58[/tex]:  

[tex]72 + 58 = 130[/tex]

Performing the Final Subtraction: Next, we subtract [tex]62[/tex] from [tex]130[/tex]:  

[tex]130 - 62 = 68[/tex]

Therefore, the final answer is [tex]68[/tex].

since they are both negative numbers just subtract 7 from 54

54-7 = 47

Answer:

The piecewise-defined function to model the salesperson's total monthly salary (in $) is,

[tex]f(x)=\begin{cases}2000 & \text{ if } x\leq 40,000 \\2000+0.05(x-40000) & \text{ if } x>40000\end{cases}[/tex]

Step-by-step explanation:

It is given that the salesperson makes a base salary of $2000 a month. Once he reaches $40,000 in total sales, he earns an additional 5% commission on the amount in sales over $40,000.

Let the x represents the amount of sales and f(x) represents the salary of salesperson.

It means till the sale of $40,000, the salary of the salesperson is constant, i.e., $2000.

[tex]f(x)=2000[/tex]    for   [tex]x\leq 40,000[/tex]

He will get commision of 5% on the amount in sales over $40,000.

[tex]f(x)=2000+\frac{5}{100}(x-40000)[/tex]  for  [tex]x>40,000[/tex]

Therefore the piecewise-defined function to model the salesperson's total monthly salary (in $) is,

[tex]f(x)=\begin{cases}2000 & \text{ if } x\leq 40,000 \\2000+0.05(x-40000) & \text{ if } x>40000\end{cases}[/tex]

The piecewise-defined function to model the salesperson's total monthly salary (in $) as a functionis [tex]\boxed{f\left( x \right)=\left\{\begin{gathered}2000{\text{ if }}x \leqslant 40000 \hfill\\2000 + 0.05\left( {x - 40000} \right){\text{ if }}x > 40000 \hfill\\\end{gathered}\right.}[/tex].

Further explanation:

The piecewise defined function is defined as a function that takes different values in different interval. It is defined on the sequence of intervals.

Given:

The base salary is [tex]\$ 2000[/tex].

He will earn [tex]5\%[/tex] additional commission on the amount sales over [tex]\$ 40000[/tex].

Explanation:

The base salary of a salesperson in a month is $2000.

If the salesperson reaches to $40000, he will earns 5% additional commission on the amount sales over [tex]\$ 40000[/tex].

The person get [tex]\$ 2000[/tex] if sale less than  [tex]\$ 40000[/tex].

[tex]f\left( x \right) =2000{\text{ for }}x \leqslant 40000[/tex].

The value of the function for less than $40000 is  [tex]f\left( x \right)=2000[/tex].

If he sales over 40000 the amount he will get can be expressed as follows,

[tex]f\left( x \right)=2000+\dfrac{5}{{100}}\left( {x - 40000} \right)[/tex] for[tex]x > 40000.[/tex]

The value of the function for greater than 40000 is  [tex]f\left( x \right)=2000 + \dfrac{5}{{100}}\left({x - 40000} \right)[/tex].

The piecewise-defined function to model the salesperson's total monthly salary (in $) as a functionis [tex]\boxed{f\left( x \right)=\left\{\begin{gathered}2000{\text{if }}x \leqslant 40000\hfill\\2000 + 0.05\left( {x - 40000}\right){\text{if }}x > 40000 \hfill \\\end{gathered}\right.}[/tex]

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear Equation

Keywords: solving equation, inverse operation, isolate, variable, addition property, good guess, solution, order of operations, reverse order, two solution, linear equation.

c=PI*d

divide both sides by PI to get d by itself

d = c/PI

To get a mixture that is 70% antifreeze, you would need 600 gallons of 80% antifreeze solution mixed with 100 gallons of 10% antifreeze solution.

To solve this problem, we will use the six step method. Let's assign variables to our unknown quantities:

Now we can set up the equation using the information given:

Simplifying the equation:

Therefore, you would need 600 gallons of the 80% antifreeze solution to mix with 100 gallons of the 10% antifreeze solution to get a mixture that is 70% antifreeze.

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Answer:   [tex]78.5\ cm^2[/tex]

Step-by-step explanation:

From the given we can see that there is a  rectangle is rotated about a line bisecting its length in segment of 5 cm in each.

Since the base will be a circle, then the area of the base will be :-

[tex]\text{Area of base}=\pi r^2\\\\\\\Rightarrow\text{Area of base}=3.14\times(5)^2\\\\\\\Rightarrow\text{Area of base}=78.5\ cm^2[/tex]

Hence, the base area of the resulting figure = [tex]78.5\ cm^2[/tex]

The approximate base area of the rectangle is rotated about a line bisecting its length in segments of 5 cm in each is 78.5.

From the given, we can see that there is a  rectangle is rotated about a line bisecting its length in segments of 5 cm in each.

Since the base will be a circle, then the area of the base will be

A=pir^2

Where r is the radius of the circle.

A=3,14(5)^2

A=78.5.

Therefore, the approximate base area of the resulting figure is 78.5.

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Rounding the coefficients to four decimal places:

[tex]\[ n(t) = 50000 \times \left(1 - e^{-20.2020t + 0.0101}\right) \][/tex]

The logistic equation governing the number of people exposed to the advertisement in the community is given by:

[tex]\[ \frac{{dn}}{{dt}} = kn \left(1 - \frac{n}{M}\right) \][/tex]

Where:

n(t)  is the number of people exposed to the advertisement at time t

k  is the growth rate constant

M  is the limiting number of people who will see the advertisement

Given:

[tex]\( n(0) = 500 \)[/tex]

[tex]\( n(1) = 1000 \)[/tex]

[tex]\( M = 50000 \)[/tex]

We first need to find the growth rate constant \( k \):

[tex]\[ 1000 = 500 \times \left(1 - \frac{500}{50000}\right) \times k \][/tex]

[tex]\[ 1000 = 500 \times \left(1 - \frac{1}{100}\right) \times k \][/tex]

[tex]\[ 1000 = 500 \times \frac{99}{100} \times k \][/tex]

[tex]\[ k = \frac{1000 \times 100}{500 \times 99} \][/tex]

[tex]\[ k = \frac{2000}{99} \][/tex]

Now, we integrate to find [tex]\( n(t) \)[/tex]:

[tex]\[ \int \frac{{dn}}{{n(1 - \frac{n}{50000})}} = \int \frac{{2000}}{{99}} dt \][/tex]

[tex]\[ -\ln|1 - \frac{n}{50000}| = \frac{{2000}}{{99}}t + C \][/tex]

Using the initial condition [tex]\( n(0) = 500 \)[/tex], we find[tex]\( C \)[/tex]:

[tex]\[ -\ln|1 - \frac{500}{50000}| = 0 + C \][/tex]

[tex]\[ -\ln\left(\frac{99}{100}\right) = C \][/tex]

So, the equation becomes:

[tex]\[ -\ln|1 - \frac{n}{50000}| = \frac{{2000}}{{99}}t - \ln\left(\frac{99}{100}\right) \][/tex]

Solving for n(t) :

[tex]\[ |1 - \frac{n}{50000}| = e^{-\frac{{2000}}{{99}}t + \ln\left(\frac{99}{100}\right)} \][/tex]

[tex]\[ 1 - \frac{n}{50000} = e^{-\frac{{2000}}{{99}}t + \ln\left(\frac{99}{100}\right)} \][/tex]

[tex]\[ \frac{n}{50000} = 1 - e^{-\frac{{2000}}{{99}}t + \ln\left(\frac{99}{100}\right)} \][/tex]

[tex]\[ n(t) = 50000 \times \left(1 - e^{-\frac{{2000}}{{99}}t + \ln\left(\frac{99}{100}\right)}\right) \][/tex]

This is the solution for [tex]\( n(t) \)[/tex]. Now, rounding the coefficients to four decimal places:

[tex]\[ n(t) = 50000 \times \left(1 - e^{-20.2020t + 0.0101}\right) \][/tex]

An equation is formed of two equal expressions. The value of y for the given equation is (1.5x - 6)/3.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

1.5x - 3y = 6

Subtract 1.5x from both the sides of the equation,

1.5x - 3y - 1.5x = 6 - 1.5x

-3y = 6 - 1.5x

Divide both the sides of the equation by -3,

-3y/(-3) = (6 - 1.5x)/(-3)

y = (1.5x - 6)/3

Hence, the value of y for the given equation is (1.5x - 6)/3.

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Answer:

A)up

Step-by-step explanation:

Given : f(x)=log x +3

            g(x)= log x

To Find :

The graph of f(x) = log x + 3 is the graph of  g(x) = log x translated 3 units

a)up

b)down

c)left

d) right

Solution:

Rule: f(x)→f(x)+b

So, the graph f(x) is translated up by b units

So, on comparing given f(x) with rule , we can say that the graph g(x) is translated up by 3 units

Thus option A is correct.

Hence The graph of f(x) = log x + 3 is the graph of  g(x) = log x translated 3 units up

To find (g-f)(3), evaluate g(3) and f(3), then subtract f(3) from g(3).

To find (g-f)(3), we need to subtract the function f(x) from g(x) and then evaluate the result at x = 3. Let's first simplify the expressions for f(x) and g(x):

f(x) = 4 - x^2

g(x) = 6x

Now substitute these into (g-f)(3) and simplify:

(g-f)(3) = g(3) - f(3)

= 6(3) - (4 - 3^2)

= 18 - (4 - 9)

= 18 - (-5)

= 23

Therefore, (g-f)(3) is equivalent to 23.

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Answer:

the answer is 6.

Step-by-step explanation:

on edge i got my question wrong and the answer was 6 <33.