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3 years ago

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When ordering tickets online, a college theater charges a $5 handling fee no matter how large the order. tickets to a comedy con

cert cost $58 each. If you had to graph the line showing the total cost, y, of buying x tickets, what would the slope of your line be? Explain your thinking.

1 answer:

Alina [70]3 years ago

6 0

Answer:

The slope is $m=\$58\ per\ ticket$

Step-by-step explanation:

Let

x ---> the number of tickets

y ---> the total cost

we know that

The equation of a line in slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-intercept

In this problem we have that

The slope or unit rate is the cost of one ticket

$m=\$58\ per\ ticket$

The y-intercept is the same that the handling fee charge

$b=\$5$

so

The linear equation that represent this situation is

$y=58x+5$

therefore

The slope is $m=\$58\ per\ ticket$

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2 years ago

During a lab experiment, the temperature of a liquid changes from 625°F to 1034°F.

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Answer:

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At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A

alexgriva [62]

Answer: A) Option B

B) 22.3 minutes.

Step-by-step explanation:

The temperature decreases in an exponential decrease, this means that we can write the equation as:

T(x) = A*r^x + B

Where A is the difference between the initial temperature and the room temperature, B is the room temperature, and r is a positive number smaller than 1, that says "how fast" the temperature decreases (and x is the variable, in units of time)

We know that the initial temperature

A = 200° - 72° = 128°

B = 72°

T(x) = 128°r^x + 72°

The only option with those two values is option B.

T(x) = 128(0.989)^x + 72

We should check that when x = 2m, the temperature must be 197°

T(2m) = 128*(0.989)^2 + 72 = 197.2° (that we can round down to 197°)

So this equation is correct

Now we want to find the time such the temperature is 172°F

then:

172° = 128°(0.989)^x + 72°

100° = 128°(0.989)^x

100/128 = (0.989)^x

x = ln(100/128)/ln(0.989) = 22.3 minutes.

Let's check in our equation:

T(22.3) = 128°(0.989)^22.3 + 72 = 172°

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3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago