what kind of tax is proposed by those who want a simple tax?
A. Progresssive Tax
B. Flat Tax
C. Tax with only 5 tax brackets
D. Those who want a simple tax want no tax

Answer:

So the questions isn't clear enough, but I could tell u that the nearest answer is the last answer, but without the multiplied 2

It's _15x_35xsquared

Answer:

5x(–3 – 7x)

(5x)(–3) + (5x)(–7x)

–15x – 35x2

Step-by-step explanation:

D

Claire's measuring stick would make a more precise measurement

The given parameters are:

Claire = 1 cm

Martin = 0.1 cm

The 0.1 cm marking would make a more accurate measurement.

This is so because the length of a slug would most likely be in decimal centimeters

However, this doesn't translate to a precise measurement.

Hence, Claire's measuring stick would make a more precise measurement

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Answer:

2

Step-by-step explanation:

Select two points that the red line goes through like (0,-10) and (2,-6).  If I start at the bottom left of the red line, I can see that I am moving to the right and up.  That means that the slope will be positive.  As my x is increasing so is my y.  Starting at point (0,-10) 2 space, but each space counts as 2, so I really move up 4.  Then I check to see how much I moved right.  It is only one space, but again, I am counting by 2s, so I really moved 2 units to the right.  

The fraction would be 4/2 which just equals 2/1  or 2.  

There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].

In this problem we must simplify the trigonometric equation by both algebraic and trigonometric means and clear the variable x:

2 · cos x = 4 · cos x · sin² x

2 · sin² x = 1

sin² x = 1/2

sin x = ± √2 /2

There are several solutions:

x₁ = π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]

x₂ = 3π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]

x₃ = 5π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]

x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]

There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].

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The original price of concert tickets before the addition of service fee is 42.93

45.50 = x + (6% of x)

45.50 = x + (0.06 × x)

45.50 = x + 0.06x

45.50 = 1.06x

x = 45.50/1.06

x = 42.9245283018867

Approximately,

x = 42.93

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The percent greater of 8 than 6 to the nearest percent is 25%

Percent greater = difference / higher chance value × 100

= 2/8 × 100

= 0.25 × 100

Percent greater = 25%

Therefore, the percent greater of 8 than 6 to the nearest percent is 25%

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The equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]

The expression is given as:

[tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex]

Divide 128 by 2

[tex]\sqrt{\frac{64x^5y^6}{x^7y^5}}[/tex]

Apply the law of indices to the variables

[tex]\sqrt{\frac{64y^{6-5}}{x^{7-5}}}[/tex]

Evaluate the differences

[tex]\sqrt{\frac{64y}{x^2}}[/tex]

Take the square root of 64

[tex]8\sqrt{\frac{y}{x^2}}[/tex]

Take the square root of x^2

[tex]\frac{8\sqrt y}{x}[/tex]

Hence, the equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]

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If these coordinates are rotated 270° counterclockwise, the resulting coordinates will be;

X' = (3, -1)

Y' = (2, -2)

Z' = (-1, -3)

If the coordinates (x, y) is rotated 270° counterclockwise, the resulting coordinate point will be;

(x, y) - > (y, -x)

From the given triangles, the coordinate points are expressed as

X = (1, 3)

Y = (2, 2)

Z = (3, -1)

If these coordinates are rotated 270° counterclockwise, the resulting coordinates will be;

X' = (3, -1)

Y' = (2, -2)

Z' = (-1, -3)

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The answers to the question are

b. (l + 2f) (w + 2f)

c. f + 4, f + 6

d.  52.25

a. The diagram is an attachment.

The equation says

(l + 2f) (w + 2f)

We have the lengths of w and l as 8 and 12 respectively. Hence we have

(8 + 2f) (12 + 2f) as the area

c. (8 + 2f) (12 + 2f)

= 96 + 16f + 24f + 4f²

= 4f² + 40 f + 96

We have to factorize this using the quadratic equation calculator

f + 4, f + 6

= f = -4 and f = -6

d. If it is 1.25 thick we would have

1.25 x 2 = 2.5

12 - 2.5 = 9.5

8 - 2.5 = 5.5

Area = l * w

= 5.5 * 9.5

= 52.25

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The key feature that the function of f(x) and g(x) has in common is the domain.

The domain of a function is the set of input or argument values for which the function is valid and well defined. The range is the set of the dependent variable for which a function is defined.

From the given information, we are to find the domain, range, x-intercept, and, y-intercept of the given equation:

[tex]\mathbf{f(x) = -4^x+5}[/tex]

[tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]

For [tex]\mathbf{f(x) = -4^x+5}[/tex];

The domain of a function [tex]\mathbf{-4^x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.

For [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]

The domain of a function [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.

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Answer:

At 15 hours of work they will be equal.

Step-by-step explanation:

James: y = 10x + 20

Kelly: y = 8X + 50

Set them equal to each other and solve

10x + 20 = 8x + 50  Subtract 8x from both sides of the equation

2x + 20 = 50  Subtract 20 from both sides of the equation

2x = 30  Divide both sides by 2

x =15

Final Answer: The inverse of f (x)=7x-4 is f^-1 (x)= (x+4)/7

Answer:

8,9,5

Step-by-step explanation:

The minimum production rate of bushels per acre of corn needed this year is 97 for two year's production to reach the goal of 49,100.

To calculate the minimum production rate for this year, we must subtract the production rate of the previous year, with what is expected to be obtained this year.

Then we must divide the minimum that must be obtained by the total area, by the 300 acres that we have, to know how much corresponds to each acre.

According to the above, each acre must produce 97 bushels of corn to reach the goal of 49,100 for both years.

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Answer:

[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]

Step-by-step explanation:

Quadratic Formula

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]

Given quadratic equation:

[tex]x^2-3x-5=0[/tex]

Define the variables:

[tex]\implies a=1, \quad b=-3, \quad c=-5[/tex]

Substitute the defined variables into the quadratic formula and solve for x:

[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)}}{2(1)}[/tex]

[tex]\implies x=\dfrac{3 \pm \sqrt{9+20}}{2}[/tex]

[tex]\implies x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]

Therefore, the exact solutions to the given quadratic equation are:

[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]

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Answer:

[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]

Step-by-step explanation:

Quadratic Formula :

[tex]\boxed {\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}}[/tex]

We are given the equation x² - 3x - 5 = 0.

Here,

Solving :

Hence, the solutions are :

[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]

Answer:

the number of 1 dollars bills was 5 si make 1 more 5 dollars so 8n total it is 10 dollars.

The solution of the formula for s is:

[tex]s = \sqrt{\frac{5V}{h}}[/tex]

The equation is:

[tex]V = \frac{1}{5}s^2h[/tex]

To solve for s, we isolate it, hence:

[tex]s^2 = \frac{5V}{h}[/tex]

[tex]s = \sqrt{\frac{5V}{h}}[/tex]

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Answer: (3, 13)

Step-by-step explanation:

If we let the point be P, then AP:BP=2:1.

[tex]P=\left(\frac{(2)(3)+(1)(3)}{2+1}, \frac{(2)(19)+(1)(1)}{2+1} \right)=(3, 13)[/tex]

Answer: A

Step-by-step explanation:

The time taken in years for money invested to accumulate to $1500 is 18.45 years

Simple interest = Total savings - principal

= 1500 - 1000

= $500

Simple interest = principal × rate × time

500 = 1000 × 0.0271 × t

500 = 27.1 × t

500 = 27.1t

t = 18.4501845018450

Approximately,

t = 18.45 years

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The [tex]n[/tex]-th term is

[tex]U_n = \dfrac14 n^2 (n+1)^2[/tex]

so the 39th term is

[tex]U_{39} = \dfrac14 39^2 40^2 = \boxed{608,400}[/tex]

Observe that

[tex]2^3 + 4^3 + 6^3 = 2^3 + 2^3\times2^3 + 2^3\times3^3 = 8\left(1^3+2^3+3^3)[/tex]

which suggests that

[tex]V_n = 8U_n = \boxed{2n^2(n+1)^2}[/tex]

The division of the figure based on the information is 14.09.

It should be noted that the question is simply to divide 860 by 61.

The division based on the information will be illustrated thus:

= 860 ÷ 61

= 14.09

In conclusion, the correct option is 14.09.

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Answer:

1777777777888876332222233

Answer:

AB / BC = 2 / 3

A equals 9 and C = 13

AC = 13 - 9 = 4

AB + BC = 13

A) AB + BC = 4

B) AB / BC = 2/3   Therefore BC = AB / (2/3)  

B) BC = 1.5 AB  

A) BC = 4 - A/B

Multiplying equation A) by -1

A) -BC = -4 + AB then we add equation B)

B) BC = 1.5 AB then adding both equations

0 = 2.5 AB -4

2.5 AB = 4

AB = 1.6

Since A = 9 then the number at B is 9 + 1.6

equals 10.6

Step-by-step explanation:

Answer:

Step-by-step explanation:

Let the number at B be x.

We know that distance between two points on the number line is the absolute value of the difference of numbers at those points.

Then distances representing the lengths of segments AB and BC are:

We are given the ratio of segments:

Substitute and solve for x:

The sum of the geometric series is 1328600

The given parameters are:

a1 = 5

r = 3

n = 12

The sum of the terms is:

[tex]S_n = \frac{a(1 - r^n)}{1 - r}[/tex]

This gives

[tex]S_n = \frac{5 * (1 - 3^{12})}{1 - 3}[/tex]

Evaluate the difference

[tex]S_n = \frac{5 * (1 - 531441)}{-2}[/tex]

This gives

[tex]S_n = \frac{5 * - 531440}{-2}[/tex]

Evaluate

[tex]S_n = 1328600[/tex]

Hence, the sum of the geometric series is 1328600

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in triangle ABC, AC=13, BC=84, and AB=85.  the measure of angle is C.5.1 degrees,

Using the The Law of Cosines in any triangle which can be expressed as;

[tex]c^2 = a^2 + b^2 - 2ab * cos(C)[/tex]

Then if we expressed the given sides in the cosine formula we have

[tex]13^2 = 85^2 + 84^2 - 2 * 85 * 84 * cos(C)[/tex]

Then we have [tex]169 = 7225 + 7056 - 14280 * cos(C)[/tex]

[tex]14280 * cosC = (7225 + 7056 - 169)[/tex]

[tex]cosC = \frac{14212}{ 14280}[/tex]

cos 0.9947

[tex]C = cos^-1(0.9947)[/tex]

C = 5.1 degrees

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Answer:

  (b)  (3,-1)

Step-by-step explanation:

One can determine the correct answer by checking to see if it satisfies the equations.

The first equation can be rewritten to ...

  x = 6 +3y . . . . . . . add 3y to both sides

This makes it easier to check answer choices:

  a) -1 = 6 +3(3) . . . . no

  b) 3 = 6 +3(-1) . . . . yes . . . . . (3, -1) is the solution

  c) 1 = 6 +3(-3) . . . . no

  d) -3 = 6 +3(1) . . . . no

__

Additional comment

We can further convince ourselves this is the correct choice by seeing if it satisfies the second equation:

  2x +2y = 4 . . . for (x, y) = (3, -1)

  2(3) +2(-1) = 4 . . . . yes

The required result based on given continuous functions is: 207. See the explanation for same below.

A continuous function in mathematics is one in which a continuous variation (that is, a change without a jump) of the argument causes a continuous variation of the function's value.

Since G and F are continuous functions,

[tex]\int_{12}^{28}) \, f(x) dx[/tex] = 19

[tex]\int_{12}^{28}) \, f(x) dx[/tex] = 35

Therefore,

[tex]\int_{12}^{28}) \, [K g(x) - 2 f(x)] dx[/tex]

= K [tex]\int_{12}^{28}) \, g(x) dx -2 \int_{12}^{28}) \, f(x) dx[/tex]

So, given that K = 7, we have

7 x 35 - 2 x 19
= 245 - 38
= 207

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I know you said "without making any assumptions," but this one is pretty important. Assuming you mean [tex]\alpha,\beta[/tex] are shape/rate parameters (as opposed to shape/scale), the PDF of [tex]X[/tex] is

[tex]f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}[/tex]

if [tex]x>0[/tex], and 0 otherwise.

The MGF of [tex]X[/tex] is given by

[tex]\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx[/tex]

Note that the integral converges only when [tex]t<2[/tex].

Define

[tex]I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx[/tex]

Integrate by parts, with

[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]

[tex]dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}[/tex]

so that

[tex]\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}[/tex]

Note that

[tex]I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}[/tex]

By substitution, we have

[tex]I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}[/tex]

and so on, down to

[tex]I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}[/tex]

The integral of interest then evaluates to

[tex]\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}[/tex]

so the MGF is

[tex]\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}[/tex]

The first moment/expectation is given by the first derivative of [tex]M_X(t)[/tex] at [tex]t=0[/tex].

[tex]\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}[/tex]

Variance is defined by

[tex]\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2[/tex]

The second moment is given by the second derivative of the MGF at [tex]t=0[/tex].

[tex]\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18[/tex]

Then the variance is

[tex]\Bbb V[X] = 18 - 4^2 = \boxed{2}[/tex]

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

[tex]M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k[/tex]

where [tex]M_X^{(k)}(0)[/tex] is the [tex]k[/tex]-derivative of the MGF evaluated at [tex]t=0[/tex]. This is also the [tex]k[/tex]-th moment of [tex]X[/tex].

Recall that for [tex]|t|<1[/tex],

[tex]\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k[/tex]

By differentiating both sides 7 times, we get

[tex]\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k[/tex]

Then the [tex]k[/tex]-th moment of [tex]X[/tex] is

[tex]M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}[/tex]

and we obtain the same results as before,

[tex]\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4[/tex]

[tex]\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18[/tex]

and the same variance follows.