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3 years ago

$6.00 times 200 per day times 65.00 per day for saftey

Mathematics

1 answer:

Molodets [167]3 years ago

4 0

78,000
................

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2 years ago

Jessica is selling books during the summer to earn money for college. She

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3 years ago

In a given year, the population of Metro City is 7420, and the population is decreasing by 152 people per year. The population o

Anit [1.1K]

Answer:

The correct option is D.

Step-by-step explanation:

Let x denote the years.

The information provided is:

The population of Metro City is 7420.
The population is decreasing by 152 people per year.
The population of Smallville is consistently 1200 less than half of Metro City.

From the provided data derive the equation for the population of Smallville x years after as follows:

$Y = [\frac{1}{2}\cdot (7420-152x)]-1200$

The population of Smallville is 2282 after x years.

Compute the value of x as follows:

$2282= [\frac{1}{2}\cdot (7420-152x)]-1200$

$\frac{7420-152x}{2}=2282+1200\\\\7420-152x=3482\times 2\\\\152x=7420-6964\\\\152x=456\\\\x=3$

Thus, the time it will take for Smallville's population to reach 2282 is 3 years.

The correct option is D.

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3 years ago

joe saved up 25 his mother gave him an additional 20 he boua movie that cost 15 how much money dose joe have left

alina1380 [7]

Answer: 30

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2 years ago

Read 2 more answers

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3>Question 1</h3>

Method 1 - modelling as a quadratic with restricted domain

Assuming that the points given on the graph are points that the curve passes through, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

Equation 1 (-4, 2)

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

Equation 2 (-1, 4)

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

Equation 3 (4, 6)

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

Method 2 - modelling as a square root function

Assuming that the points given on the graph are points that the curve passes through, and the x-intercept should be included, we can model this curve as a square root function.

Given points: