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3 years ago

In a 2560 gallons there are 1944gallons left after two weeks how many days a whole tank will last

1 answer:

6 0

In this problem all the necessary informations are provided. Total capacity of the tank is given as 2560 gallons. It is also given that after 2 weeks or 14 days , only 1944 gallons of water is left. The number of days the whole tank will be empty is required to be found.
Let us assume that the number of days required to empty the tank = x
Then
Now amount of water consumed in 14 days = 2560 - 1944
= 616 gallons
So
616 gallons of water used in = 14 days
2560 gallons of water will be used in = (14/616) * 2560
= 58.18
Then the whole tank will last approximately 58 days if calculated to the nearest whole number.

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Step-by-step explanation:

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3 years ago

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: