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2 years ago

Write the quadratic equation whose roots are -3 and 3, and whose leading coefficient is 5

Mathematics

1 answer:

anastassius [24]2 years ago

3 0

Answer:

5x²- 1.8

Step-by-step explanation:

roots are -3 and 3, put them in brackets

(x - 3)(x + 3)

x² + 3x - 3x - 9

x² - 9

5x = -9

x = -1.8

5x² - 1.8

just to check:

5x² - 1.8

x² - 9

(x - 3)(x - 3)

x = -3 and 3.

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A cylindrical swimming pool has a diameter of 20 feet and a height of 3.5 feet. How many gallons of water can the pool contain?

kvasek [131]

Answer:

The swimming pool will contain approx $8243\ gal$ of water.

Step-by-step explanation:

Given the swimming pool is in a cylindrical shape.

And the dimensions are $20\ feet$ diameter with a height of $3.5\ feet.$

Also, $1\ ft^3 =7.5\ gal$

First, we will find the volume of the swimming pool.

The swimming pool is in cylindrical shape.

So, the volume would be $\pi r^2h$

The height $(h)$ is given as $3.5\ feet.$ To calculate radius $(r)$ we will divide the diameter by two.

The radius of the cylinder would be $\frac{20}{2}=10\ feet$

Plug these values in the formula we get,

$\pi r^2h\\=\pi\times 10^2\times 3.5\\=3.14\times 100\times 3.5\\=1099\ ft^3$

Now, it is given $1\ ft^3 =7.5\ gal$

So, $1099\ ft^3 =1099\times 7.5\ gal=8242.5\ gal$

The swimming pool will contain approx $8243\ gal$ of water.

6 0

2 years ago

Which situation is linear ?

QveST [7]

Answer:

D. The population of a town in Increasing by ten people each year

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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

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

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