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3 years ago

Fundamental Theorem of Algebra...

Mathematics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

A real root of fifth-grade multiplicity/No complex roots.

Step-by-step explanation:

The Fundamental Theorem of Algebra states that every polynomial with real coefficients and a grade greater than zero has at least a real root. Let be $f(x) = (x+7)^{5}$ , if such expression is equalized to zero and handled algebraically:

1) $(x+7)^{5} = 0$ Given.

2) $(x+7)\cdot (x+7)\cdot (x+7)\cdot (x+7)\cdot (x+7) = 0$ Definition of power.

3) $x+7=0$ Given.

4) $x = -7$ Compatibility with the addition/Existence of the additive inverse/Modulative property/Result.

This expression has a real root of fifth-grade multiplicity. No complex roots.

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Which term best completes the state all _______ are rectangles A. Parallelograms B.squares C.rhombi D. None of theses

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Answer:

The answer is B. squares.

Step-by-step explanation:

All squares are considered rectangles. Not all parallelograms are rectangles, and rhombi are not rectangles.

I hope this helped! :-)

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3 years ago

Isosceles triangles have two equal sides. true or false

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True
isosceles triangles
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7 1/2- 3 7/10 or Seven and a half subtract three and seven tenths.

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3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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3 years ago

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Airida [17]

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