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3 years ago

3 folders cost $2.91

Mathematics

2 answers:

Arisa [49]3 years ago

6 0

Answer:

2/×=3/$2.91

Step-by-step explanation:

WARRIOR [948]3 years ago

3 0

Answer: E. None of the above

Step-by-step explanation:

The equation to solve the problem should look like...2x = $2.91 / 3

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What is the relationship between the factors and the graph of a polynomial equation??

bija089 [108]

Answer:

Each intercept corresponds to a zero of the polynomial function and every zero yields a factor.

Step-by-step explanation:

Each intercept corresponds to a zero of the polynomial function and every zero yields a factor, so we can now write the polynomial in factored form. Graphs behave differently at various intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off.

8 0

2 years ago

A is true but f(x) is wrong need help

MrRissso [65]

Answer:

f(x)=-18x^2

Step-by-step explanation:

Given:

1+Integral(f(t)/t^6, t=a..x)=6x^-3

Let's get rid of integral by differentiating both sides.

Using fundamental of calculus and power rule(integration):

0+f(x)/x^6=-18x^-4

Additive Identity property applied:

f(x)/x^6=-18x^-4

Multiply both sides by x^6:

f(x)=-18x^-4×x^6

Power rule (exponents) applied"

f(x)=-18x^2

Check:

1+Integral(-18t^2/t^6, t=a..x)=6x^-3

1+Integral(-18t^-4, t=a..x)=6x^-3

1+(-18t^-3/-3, t=a..x)=6x^-3

1+(6t^-3, t=a..x)=6x^-3

That looks great since those powers are the same on both side after integration.

Plug in limits:

1+(6x^-3-6a^-3)=6x^-3

We need 1-6a^-3=0 so that the equation holds true for all x.

Subtract 1 on both sides:

-6a^-3=-1

Divide both sides by-6:

a^-3=1/6

Raise both sides to -1/3 power:

a=(1/6)^(-1/3)

Negative exponent just refers to reciprocal of our base:

a=6^(1/3)

3 0

3 years ago

Simplify. Assume that a is a positive integer and a>2. (a-2)!/a!

Paul [167]

Answer:

C. 1 / (a(a - 1))

Step-by-step explanation:

View Image

Just know that:

n! = n(n-1)!

= n(n-1)(n-2)!

= n(n-1)(n-2)(n-3)!

= ...

8 0

3 years ago

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Hi please help me thanksss<br>pls provide workings too thanks :)

anygoal [31]

Step-by-step explanation:

Search for questions & chapters

Class 11

>>Maths

>>Trigonometric Functions

>>Trigonometric Equations

>>Solve for x: sin x + sin 2x...

Question

Bookmark

Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.

Medium

Solution

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We write the given equation as

(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x

or 2sin2xcosx+sin2x=2cos2xcosx+cos2x

or sin2x(2cosx+1)=cos2x(2cosx+1)

or (sin2x−cos2x)(2cosx+1)=0

∴sin2x−cos2x=0 or 2cosx+1=0

If sin2x−cos2x=0, then tan2x=1,

Hence 2x=nπ+π/4

or x=(4n+1)

.(1)

If 2cosx+1=0, then cosx=−1/2 (2)

∴x=2nπ±

2π

or x=

6n±2

We seek values of x in the interval 0≤x≤2π

In this interval (1) gives

x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)

and (2) gives x=2π/3,4π/3. (for n=0,1)

Thus we get the answer

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

4 0

2 years ago

Write the following statement as expression 50 more than x

spin [16.1K]

Answer:

What statement??? I don't see one to re-write.

Step-by-step explanation:

5 0

2 years ago

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