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3 years ago

Which expression makes the equation true?

1 answer:

N76 [4]3 years ago

8 0

1. C.

The rule of powers is that when a parenthesis is raised to a power, you multiply the two numbers. Since it is being raised to the second power, it needs to have a 10 in the parenthesis. That would make 20.

2. B

Firstly 10^2 = 100, so we know that the number must be 10. That leaves us with just A and B. Then we know the same thing from the previous equation, that they need to be multiplied together.

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Create a 3rd degree polynomial function with one zero at three. Sketch a graph

DedPeter [7]

Answer:

$f(x) = {x}^{3} - 9 {x}^{2} + 27x - 27$

Step-by-step explanation:

We want to create a third degree polynomial function with one zero at three.

In other words, we want to find a polynomial function with roots x=3 , multiplicity, 3.

Since x=3 is a solution, x-3 is the only factor that repeats thrice.

$f(x) = {(x - 3)}^{3}$

We expand to get:

$f(x) = x( {x}^{2} - 6x + 9) - 3( {x}^{2} - 6x + 9)$

$f(x) = {x}^{3} - 6 {x}^{2} + 9x- 3 {x}^{2} + 18x - 27$

This simplifies to:

$f(x) = {x}^{3} - 9 {x}^{2} + 27x - 27$

See attachment for graph.

4 0

3 years ago

A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost

Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

S = 2πr² + 2πrh ( surface area of the lids + surface are of the curved surface)

Now the function that represents the cost to construct the can will be

C = 2πr²(0.027) + 2πrh(0.015)

C = 0.054πr² + 0.03πrh ---------(1)

Volume of the can = Volume of a cylinder = πr²h

16 = πr²h

$h=\frac{16}{\pi r^{2}}$ -------(2)

Now we place the value of h in the equation (1) from equation (2)

$C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})$

$C=0.054\pi r^{2}+0.03(\frac{16}{r})$

$C=0.054\pi r^{2}+(\frac{0.48}{r})$

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

$C'=0.108\pi r-(\frac{0.48}{r^{2} })$

Now for C' = 0

$0.108\pi r-(\frac{0.48}{r^{2} })=0$

$0.108\pi r=(\frac{0.48}{r^{2} })$

$r^{3}=\frac{0.48}{0.108\pi }$

r³ = 1.415

r = 1.12 inch

and h = $\frac{16}{\pi (1.12)^{2}}$

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

$C"=0.108+\frac{0.48}{r^{3}}$ which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

5 0

3 years ago

Math:

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

$x = \cos^{-1} \left(-\frac{1}{2} \right)$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

7 0

3 years ago

Can someone tell me what to write of how i found the answer please ?!

Vlad [161]

Answer:

108

Step-by-step explanation:

4−22+135−9

=−18+135−9

=117−9

=108

5 0

3 years ago

Describe how both the Rational Root Theorem and Descartes’ Rules of Signs help you to find the zeros of a polynomial? Give me an

MrRa [10]

Answer:

Step-by-step explanation:

Rational Root Theorem: If the polynomial

P(x) = a n x n + a n – 1 x n – 1 + ... + a 2 x 2 + a 1 x + a 0

has any rational roots, then they must be of the form of (factors of a0/factors of an).

Example: F(x) = 4x² + 5x +2

If this polynomial has any rational roots, then they must be (factors of 2)/(factors of 4), so (±1, ±2)/(±1, ±2, ±4). So if this polynomial has any rational roots, they must be either: ±1, ±1/2, ±1/4, or ±2. Notice that this polynomial doesn't have to have any rational roots, but if it does, then the roots must fit the Rational Root Theorem.

Descartes' Rules of Signs:

a). In a polynomial, how many time the sign changes is how many positive roots the polynomial will have.

Example: 5x³ + 6x² - 2x - 1

In this expression, the sign only changed once, between 6x² and 2x, so it will only have one positive root.

Example 2: 6x³ - 4x² + x - 6

In this expression, the sign changed 3 times (remember there is a invisible "+" sign before the 6x³), so it will have 3 positive roots.

b). In a polynomial, if you plug in "-x" for all "x", then how many times the new polynomial changes sign is how many negative roots the old polynomial have.

Example: 5x³ + 6x² - 2x - 1.

If we plug in "-x" for all "x", then we get 5(-x)³ + 6(-x)² - 2(-x) -1, which simplifies to -5x³ + 6x² + 2x -1. In this new expression, the sign changed twice, so we have two negative roots for the expression. Notice how we got one positive root the first time and two negative roots the second time, and 1 + 2 = 3. The Fundamental Theorem of Algebra states that for a nth degree polynomial, it will have n complex roots. The polynomial we worked with was a 3rd degree polynomial, and we got 1 + 2 = 3 roots in the end.

4 0

3 years ago