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3 years ago

Find the zeros of the function g(x) = 2x2 - 5x + 2.

1 answer:

8 0

G(x) = 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x · x - 2x · 2 - 1 · x - 1 · (-2)

= 2x(x - 2) -1(x - 2) = (x - 2)(2x - 1)

g(x) = 0 ⇔ (x - 2)(2x - 1) = 0 ⇔ x - 2 = 0 or 2x - 1 = 0

x = 2 or x = 0.5

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3 years ago

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Step-by-step explanation:

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2 years ago

Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C

Lesechka [4]

Answer:

Solution : $-\frac{3}{4}-\frac{3}{4}i$

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$

= $-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)$ ÷ $2\sqrt{2}\left(0-1\right)i$

= $3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right)$ ÷ $-2\sqrt{2}i$

= $\frac{3\left(1-i\right)}{\sqrt{2}}$ ÷ $2\sqrt{2}i$ = $-3-3i$ ÷ $4$ = $-\frac{3}{4}-\frac{3}{4}i$

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2 years ago

Which of the following can be used to calculate the volume of water inside the fish bowl?

Lunna [17]

The volume of a sphere can be calculated as

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

Where r is the radius of the sphere

We want to calculate half of the volume, then we must divide that volume by 2

$V^{\prime}=\frac{1}{2}\frac{4}{3}\pi r^3$

Now we must find the radius of our sphere, the segment AB is the diameter of the sphere, and the radius is half od the diameter, then

$r=\frac{AB}{2}=\frac{12}{2}=6$

Let's put it into our equation

$\begin{gathered} V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(r)^3 \\ \\ V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(6)^3 \end{gathered}$

The problem says to use

$\pi\approx\frac{22}{7}$

Then

$V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

Final answer:

The formula that can be used to calculate the volume of water inside the fish bowl is

$V=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

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1 year ago