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3 years ago

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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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

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

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