The rectangular poster shown measures 2w-30 by w. The poster has an area of 5400 cm2. What is the value of w?

Answer:

Step-by-step explanation:

We have given:

x^4-2/x+1

Put x+1 = 0

x= 0-1

x= -1

-1 | 1   0   0   0   -2

  |      -1   1   -1     1

_______________

    1    -1   1  -1    -1

Thus it makes the expression:

x^3-x^2+x-1 - 1/x+1

You can further confirm this expression by solving the expression:

(x4 − 2) ÷ (x + 1).

x^4 - 1 -1/x+1

x^4-1/x+1 - 1/x+1

(x^2-1) (x^2+1)/x+1 - 1/x+1

(x-1)(x+1) (x^2+1)/x+1  - 1/x+1

x+1 n the numerator will be cancelled out by x+1 in the denominator.

(x-1) (x^2+1) - 1/x+1

Multiply (x^2+1) by x-1

x(x^2+1) -1(x^2+1)  - 1/x+1

x^3+x-x^2-1 - 1/x+1

x^3-x^2+x-1 - 1/x+1 ....

Sum of two monomials is not necessarily always a monomial.

For example:

Suppose we have two monomials as 2x and 5x.

Adding 2x+5x , we get 7x.

So if two monomials are both like terms then their sum will be a monomial.

Suppose we have two monomials as 3y and 4x

Now these are both monomials but unlike, so we cannot add them together and sum would be 3y + 4x , which is a binomial.

So if we have like terms then the sum is monomial but if we have unlike terms sum is binomial.

Product of monomials:

suppose we have 2x and 5y,

Product : 2x*5y = 10xy ( which is a monomial)

So yes product of two monomials is always a monomial.

The sum of two monomials is not always a monomial and the product of two monomials is always a monomial.

Further explanation:

Explanation:

Consider the two monomials as [tex]5y{\text{ and }}7y.[/tex]

The sum of the two polynomials can be obtained as follows,

[tex]\begin{aligned}{\text{Sum}} &= 5y + 7y\\&= 12y\\\end{aligned}[/tex]

The sum of the two monomials is monomial.

Consider the two monomials as [tex]5y{\text{ and }}7x.[/tex]

The sum of the two polynomials can be obtained as follows,

[tex]{\text{Sum}} = 5y + 7x[/tex]

The sum of the two monomials is not monomial.

Consider the two monomials as [tex]4y{\text{ and }}6x.[/tex]

The product of the two polynomials can be obtained as follows,

[tex]\begin{aligned}{\text{Product}} &= 4y \times 6x\\&= 24xy\\\end{aligned}[/tex]

The product of the two monomials is monomial.

The sum of two monomials is not always a monomial and the product of two monomials is always a monomial.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Number system

Keywords: sum, two monomial, monomials, always monomial, product, always, monomial, factor, factorization, polynomial, quadratic, cubic, greatest common factor, groups, multiplication, product, identities, common factor, expression, terms, grouping.

Answer:

the answer is V≈19.68cm³

V=a3

a

Edge

cm

Step-by-step explanation:

Answer:

18

Step-by-step explanation:

[tex]90[/tex] ÷ [tex]\frac{250}{3} = 1.08[/tex]

This is 1% of the trip

[tex]1.08[/tex] · [tex]100 = 108[/tex]

This is the amount of the whole trip

[tex]108 - 90 = 18[/tex]

So the answer is he has to 18 more miles to travel.

Given is the damage index equation, [tex] y=\frac{-2}{3} x+14 [/tex]

where y represents damage index, and x represents ambient temperature.

It is similar to slope-intercept form of linear equation given by y = mx + b.

Where m is said to be the slope of line and b is the y-intercept of the line. When x = 0, y = b.

Similarly, if plug x = 0 in the damage index equation, [tex] y=\frac{-2}{3} (0)+14=14 [/tex]

So, If ambient temperature, x = 0. Then damage index, y = 14.

Hence, option D is correct i.e. " If the ambient temperature is 0 then the damage index would be 14".

Answer:

The answer is D because 14 is the y-intercepts, which means that the x-value is 0. So the correct meaning is If the ambient temperature is 0 then the damage index would be 14. .

Step-by-step explanation:

Hence, the mid-point of side EF is located at:

                    (-0.5,4.5)

We know that the coordinates of a mid-point C(e,f) of a line segment AB with vertices A(a,b) and B(c,d) is given by:

[tex]e=\dfrac{a+c}{2},\ f=\dfrac{b+d}{2}[/tex]

Here we have to find the mid-point of side EF.

E(-2,3) i.e. (a,b)=(2,3)

and F(1,6) i.e. (c,d)=(1,6)

Hence, the coordinate of midpoint of EF is:

  [tex]e=\dfrac{-2+1}{2}\ ,\ f=\dfrac{3+6}{2}\\\\\\e=\dfrac{-1}{2}\ ,\ f=\dfrac{9}{2}\\\\\\e=-0.5\ ,\ f=4.5[/tex]

                Hence, the mid-point of side EF is:

                          (-0.5,4.5)

Answer:

[tex]Arc length=32.97m[/tex]        

Step-by-step explanation:

It is given that In a circle with a radius of 6 m, an arc is intercepted by a central angle of [tex]\frac{7\pi}{4}[/tex] radians.

Then, the arclength of the circle is given as:

[tex]Arc length=radius{\times}central angle[/tex]

⇒[tex]Arc length=6{\times}\frac{7\pi}{4}[/tex]

⇒[tex]Arc length=\frac{21\pi}{2}[/tex]

⇒[tex]Arc length=10.5{\times}3.14[/tex]

⇒[tex]Arc length=32.97m[/tex]

Thus, the arc length of the given circle is 32.97m.

The solution of the equation [tex]-x^2+2x+3=x^2-2x+3[/tex] are the x-coordinates of common points on the graphs of the function [tex]y=-x^2+2x+3[/tex] and [tex]y=x^2-2x+3.[/tex]

As you can see from the attached diagram these graphs intersect at points (0,3) and (2,3), then the solutions of the previous equation are x=0 and x=2.

Answer: x=0, x=2.

The best value for the interval width is 17

The highest and the lowest data elements are given as:

x = 96

x = 11

Start by calculating the range

Range = High - Low

So, we have:

Range = 96 - 11

Range = 85

The interval width depends on the number of classes.

Assume that the number of classes is 5,

Then the width is calculated as:

Width = Range / Class

So, we have:

Width = 85/5

Width = 17

Hence, the number of interval width is 17

Read more about interval width at:

https://brainly.com/question/16631975