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2 years ago

Which algebraic expresion is a polynomial with a degree of 2

Mathematics

1 answer:

MariettaO [177]2 years ago

3 0

A polynomial with a degree of 2 can be:

x² + 1

y² + 2

etc..

hope this helps

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Then solve each problen

marysya [2.9K]

Answer:

a(0.80)=60

a=60/0.80

a=75

8 0

2 years ago

Frances bought 144 shares of dwyn horticulture for $7.73 apiece. her broker charged her a commission of $46.15 for the purchase.

Svetach [21]

The annual yield on Frances’s stock approximately will be 0.0758. Then the correct option is A.

<h3>What is Algebra?</h3>

Algebra is the study of mathematical symbols and the rule involves manipulating these mathematical symbols.

Frances bought 144 shares of Dwyn horticulture for $7.73 apiece. her broker charged her a commission of $46.15 for the purchase.

Then the total amount spent by the company will be

→ 144 × $7.73 + $46.15 = $ 1159.27

If the yearly dividend on Dwyn horticulture is 61 cents per share, then the annual yield on Frances’s stock will be

$\Rightarrow \dfrac{0.61 \times 144}{11592.27} \\\\\Rightarrow 0.07577$

More about the Algebra link is given below.

brainly.com/question/953809

3 0

2 years ago

Read 2 more answers

Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2

Anastaziya [24]

Answer:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

Step-by-step explanation:

we would like to figure out the derivative of the following:

$\displaystyle \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

to do so, let,

$\displaystyle y = \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

By simplifying we acquire:

$\displaystyle y = 3 - \frac{2}{x} - \frac{1}{ {x}^{2} }$

use law of exponent which yields:

$\displaystyle y = 3 - 2 {x}^{ - 1} - { {x}^{ - 2} }$

take derivative in both sides:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )$

use sum derivation rule which yields:

$\rm\displaystyle \frac{dy}{dx} = \frac{d}{dx} 3 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

By constant derivation we acquire:

$\rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

use exponent rule of derivation which yields:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )$

simplify exponent:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )$

two negatives make positive so,

$\displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3}$

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

$\displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3}$

simplify addition:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

and we are done!

5 0

2 years ago

18. At a putt-putt course there are 50 yellow golf balls, 45 red golf

TEA [102]

Here is the answer to you question

5 0

2 years ago

How does the graph y=3^x compare to the graph y=3^-x

LUCKY_DIMON [66]

Answer:

Y = 3x^x is a graph that has exponential growth while y = 3^-x has exponential decay.

Y = 3x^x (-∞, 0) and (∞, ∞).

Y = 3x^-x (-∞, ∞) and (∞, 0).

Step-by-step explanation:

The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".

When graph 1 had positive ∞ for its x value, its y value was reaching towards positive ∞. When its x was reaching for negative ∞, its y was going for 0.

For graph 2, however, when its x was reaching for positive ∞, its x was reaching for 0. When its x was reaching for negative ∞, its y was going for positive ∞.

Here's an image of the graphs:

5 0

2 years ago