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2 years ago

Classify the polynomial 5m^10

Mathematics

1 answer:

Naily [24]2 years ago

3 0

Answer:

Monomial

Step-by-step explanation:

A polynomial has terms in their equation.

When it is a trinomial, it usually follows the pattern such as x^2+2x+1.

When it is a binomial, it usually is something like x+2

...And finally, when it is a monomial, it can be anything that has just one term. So let's say 4x^8883293483.

The polynomial 5m^10 classifies under a monomial since it only has one term

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Graph the function f(x) = x3 – 5x2 + 6x using the graphing calculator. What are the solutions to the related equation? Check all

Juli2301 [7.4K]

the solutions to the related equation are 0,2,3 .

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

Here we have , function f(x) = x3 – 5x2 + 6x . Graph of this function is given below . We need to find What are the solutions to the related equation . Let's find out:

Solution of graph means the value of x at which the value of f(x) or function is zero . We can determine this by seeing the graph as at what value of x does the graph intersect or cut x-axis !

At x = 0 .

From the graph , at x=0 we have f(x) = 0 i.e.

⇒ $f(x) = x^3- 5x^2 + 6x$

⇒ $f(0) = 0^3- 5(0)^2 + 6(0)$

⇒ $f(0) =0$

At x = 2 .

From the graph , at x=2 we have f(x) = 0 i.e.

⇒ $f(x) = x^3- 5x^2 + 6x$

⇒ $f(0) = 2^3- 5(2)^2 + 6(2)$

⇒ $f(0) =0$

At x=3 .

From the graph , at x=3 we have f(x) = 0 i.e.

⇒ $f(x) = x^3- 5x^2 + 6x$

⇒ $f(0) = 3^3- 5(3)^2 + 6(3)$

⇒ $f(0) =0$

Therefore , the solutions to the related equation are 0,2,3 .

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3 years ago

A restaurant sells about 330 sandwiches each day at a price of $6 each $.25 decrease in price. 15 more sandwiches are sold per d

zmey [24]

Answer:

restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.

the revenue is $1983.75

Step-by-step explanation:

to calculate current revenue= $6 x 330 = $1980

suppose x as the number of times the price to be dropped by $0.25

then find new price.. i.e

new price= $(6-0.25x)

and, new sell=330 +15x sandwiches

therefore, the new revenue would be= (6-0.25x)(330 +15x)

in order to maximize the current revenue, simplify the above equation and make it complete square using x

(6-0.25x)(330 +15x)

=1980-82.5x +90x -3.75 $x^{2}$

=1980 + 7.5x -3.75 $x^{2}$

=1980-3.75 (-2x+ $x^{2}$ ) ----> taking out common

now, to make a complete square lets add and subtract 1 inside the parentheses

=1980-3.75(-1+1-2x+ $x^{2}$ )

=1980 +3.75 -3.75( $x^{2}$ -2x +1)

=1983.75 -3.75 $(x-1)^{2}$ ---->(1)

as $(x-1)^{2}$ is positive always, minimize the other term in order to maximize the total revenue.

so the minimum possible value of $(x-1)^{2}$ = 0

therefore, x=1

putting x in eq(1) the revenure becomes,

$(1983.75-0)=> $1983.75

therefore, restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.

the revenue is $1983.75

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3 years ago

A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that

PilotLPTM [1.2K]

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. The height of the mountain is 3110.5767 feet.

<h3>What is Tangent (Tanθ)?</h3>

The tangent or tanθ in a right-angle triangle is the ratio of its perpendicular to its base. it is given as,

$\rm Tangent(\theta) = \dfrac{Perpendicular}{Base}$