Which of the following is a quadratic function?

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dlinn [17]

The answer is B.24x^6-18x^5=6x^5(4x-3)

3 0

3 years ago

Identify the excluded values of this product. Then rewrite the product in simplest form.

lara [203]

$\frac{6 {y}^{2} + 18y - 60 }{3 {y}^{2} - 12y } \times \frac{ {y}^{2} - 16 }{ {y}^{2} + 2y - 8} \\ \\ \frac{ {6y + 30y - 12y - 60}^{2} }{3y(y - 4)} \times \frac{ {(y)}^{2} - ( {4)}^{2} }{ {y}^{2} + 4y - 2y - 8 } \\ \\ \frac{6y(y + 5) - 12(y + 5)}{3y(y - 4)} \times \frac{(y + 4)(y - 4)}{y(y + 4) - 2(y + 4)} \\ \\ \frac{(6y - 12)(y + 5)}{3y(y - 4)} \times \frac{(y + 4)(y - 4)}{(y + 4)(y - 2)} \\ \\ \frac{6(y - 2)(y + 5)}{3y(y - 2)} \\ \\ \frac{2(y + 5)}{y} .$

8 0

1 year ago

Please help! 30 points!

Rashid [163]

Answer: $64.2^{\circ}$

Step-by-step explanation:

According to the cosine rule

$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Here, $a=83, b=111, c=165,\ C=180^{\circ}-\theta$

Putting values

$\Rightarrow \cos (180^{\circ}-\theta)=\dfrac{83^2+111^2-165^2}{2\times 83\times 111}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{6889+12321-27225}{18426}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{-8015}{18426}=-0.4349\\\\\Rightarrow 180^{\circ}-\theta=\cos^{-1}(-0.4349)=115.77^{\circ}\\\Rightarrow \theta=64.23\approx 64.2^{\circ}$

So, the ship turns by $64.2^{\circ}$ north of west

8 0

3 years ago

A storage tank will have a circular base of radius r and a height of r. The tank can be either cylindrical or hemispherical (ha

Neko [114]

Answer:

Volume: $\frac{2}{3}\pi r^3$

Ratio: $\frac{2}{9}r$

Step-by-step explanation:

First of all, we need to find the volume of the hemispherical tank.

The volume of a sphere is given by:

$V=\frac{4}{3}\pi r^3$

where

r is the radius of the sphere

V is the volume

Here, we have a hemispherical tank: a hemisphere is exactly a sphere cut in a half, so its volume is half that of the sphere:

$V'=\frac{V}{2}=\frac{\frac{4}{3}\pi r^3}{2}=\frac{2}{3}\pi r^3$

Now we want to find the ratio between the volume of the hemisphere and its surface area.

The surface area of a sphere is

$A=4 \pi r^2$

For a hemisphere, the area of the curved part of the surface is therefore half of this value, so $2\pi r^2$ . Moreover, we have to add the surface of the base, which is $\pi r^2$ . So the total surface area of the hemispherical tank is