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kotegsom [21]
3 years ago
9

Find the solution to the system of equations. Select all that apply. Y=x^2-2x-3. Y=x-3

Mathematics
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

The solution to the system of equations are (0 , -3) ⇒ answer B and

(3 , 0) ⇒ answer D

Step-by-step explanation:

* Lets explain the problem

- The quadratic equation represented by parabola

- The linear equation represented by line

- If the two equation solved together, then we have 3 types of solutions

# They intersected in two different points (2 solution)

# They intersected in one point (1 solution)

# They did not intersect in any points (No solution)

* Now lets solve the problem

∵ The quadratic equation is y = x² - 2x - 3

- It represented by parabola (red graph)

∵ The linear equation is y = x - 3

- It represented by line (blue graph)

- The two graphs intersect each other at points (3 , 0) and (0 , -3)

∵ The solution of the system of the equation is the intersection points

  between the two graphs

∵ The two graphs intersect each other at points (3 , 0) and (0 , -3)

∴ The solution to the system of equations are (3 , 0) and (0 , -3)

* The solutions are B and D

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Answer:

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Step-by-step explanation:

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5 0
2 years ago
. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum
Drupady [299]

Answer:

r=1.05\ \text{ft}

h=2.12\ \text{ft}

20.91\ \text{ft}^3

Step-by-step explanation:

r = Radius

h = Height

Volume of cylinder = 7.35\ \text{ft}^3

V=\pi r^2h\\\Rightarrow h=\dfrac{V}{\pi r^2}\\\Rightarrow h=\dfrac{7.35}{\pi r^2}

Surface area is given by

A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r^2+2\pi r\dfrac{7.35}{\pi r^2}\\\Rightarrow A=2\pi r^2+\dfrac{14.7}{r}

Differentiating with respect to radius we get

\dfrac{dA}{dr}=4\pi r-\dfrac{14.7}{r^2}

Equating with zero we get

0=4\pi r-\dfrac{14.7}{r^2}\\\Rightarrow 4\pi r=\dfrac{14.7}{r^2}\\\Rightarrow r^3=\dfrac{14.7}{4\pi}\\\Rightarrow r=(\dfrac{14.7}{4\pi})^{\dfrac{1}{3}}\\\Rightarrow r=1.05

\dfrac{d^2A}{dr^2}=4\pi-2\times \dfrac{14.7}{r^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=4\pi+2\times \dfrac{14.7}{1.05^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=37.96>0

So, the value of the function is minimum at r=1.05

h=\dfrac{7.35}{\pi r^2}=\dfrac{7.35}{\pi 1.05^2}\\\Rightarrow h=2.12

So, the radius and height which would minimize the surface area is 1.05 feet and 2.12 feet respectively.

Surface area

A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi \times 1.05^2+2\pi 1.05\times 2.12\\\Rightarrow A=20.91\ \text{ft}^3

The minimum surface area is 20.91\ \text{ft}^3.

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Step-by-step explanation:

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