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Elden [556K]
3 years ago
9

Let f(x) = x - 1/x. which of the following are zeros of f(x)?

Mathematics
1 answer:
ZanzabumX [31]3 years ago
7 0
I'm assuming that is f(x) = (x-1)/x
In that case, the answer would be just x = 1.

However, if the original is f(x) = x - (1/x) the answer would be x = 1 and x = -1.

It's really just a formatting thing—if the given equation is indeed f(x) = (x-1)/x then x = 1 is the answer. 

Hope this helped!! xx
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Solve for x to the nearest tenth...please help me lol
sleet_krkn [62]

Answer:

<h2><u>x = 8.2</u></h2>

Step-by-step explanation:

A^2+B^2=C^2 is the pythagorean theorem to find a missing siede of a triangle.

First triangle on the right:

5^2+B^2=10^2

25+B^2=100

B^2=75

B=8.66025404

B=8.7

Second triangle of left:

3^2+B^2=8.7^2

9+B^2=75.69

B^2=66.69

B= 8.16639455

B=8.2

x=8.2

3 0
3 years ago
Which is bigger -7/11 or -2/3
Bad White [126]

Answer:

-2/3

Step-by-step explanation:

3 0
3 years ago
GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial vel
zhuklara [117]
<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

h(t) = -16t^2 + 60t

<em><u>At what times will the object be at a height of 56 feet</u></em>

<em><u>Substitute h = 56</u></em>

56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0

Solve the above equation by quadratic formula

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

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