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3 years ago

14

X^2-2x−4=0 please leave ans in surd form and simplify

2 answers:

Vlada [557]3 years ago

6 0

<em>Hi there!</em>

<em>How are you?</em>

<em>Answer:</em>

<em /> $\boxed{x = + \sqrt{5} |Or| x = 1 - \sqrt{5} }$ <em />

<em>Step-by-step explanation:</em>

<em>First you have to use quadratic formula with </em> $a=1, b=-2, c=-4.$ <em />

<em /> $x = \frac{-b \± \sqrt{b^2 -4ac} }{2a}$ <em />

<em /> $x = \frac{-(-2) \± \sqrt{(-2)^2 - 4(1)(-4)} }{2(1)}$ <em />

<em /> $x = \frac{\sqrt{^2\± \sqrt{20} } }{2}$ <em />

<em /> $x = + \sqrt{5}$ <em> or </em> $x = 1 - \sqrt{5}$ <em />

<em>Hvae a good day/night!</em>

Jobisdone [24]3 years ago

3 0

Step-by-step explanation:

using the quadratic formula......

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Taya2010 [7]

Answer:

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Step-by-step explanation:

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3 years ago

What is the slope of a line perpendicular to the line whose equation is 9x-6y=36

VladimirAG [237]

Answer: -2/3

Step-by-step explanation:

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3 years ago

Read 2 more answers

. Each of the following has 9 as a factor except<br> A) 18018 B) 18 081 C) 81 180<br> D) 81 181

Arlecino [84]

Answer:

d

Step-by-step explanation:

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4 0

3 years ago

Given h(x) = 3x - 3, find h(3).

Daniel [21]

Answer:

<h2>h(3) = 6</h2>

Step-by-step explanation:

h(x) = 3x - 3

To find h(3) , substitute the value of x that's 3 into h(x). That is for every x in f (x) replace it with 3

We have

h(3) = 3(3) - 3

= 9 - 3

We have the final answer as

<h3>h(3) = 6</h3>

Hope this helps you

4 0

3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

3 years ago