Write and solve an equation to find the value of x.

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]

Please see the blue curve of the image attached below to know the graph of the function g(x) = (1/3) · 2ˣ.

Herein we have an original function f(x). The transformed function g(x) is the result of compressing f(x) by 1/3. Then, we find that g(x) = (1/3) · 2ˣ. Lastly, we graph both function on a Cartesian plane with the help of a graphing tool.

The result is attached below. Please notice that the original function f(x) is represented by the red curve, while the transformed function g(x) is represented by the blue curve.

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The expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]

From the question, we are to determine the expression for a in terms of c

In the given right triangle, we can write that

[tex]c^{2} = a^{2} +3^{2}[/tex] (Pythagorean theorem)

Thus,

[tex]c^{2} = a^{2} +9[/tex]

[tex]a^{2}= c^{2} -9[/tex]

[tex]a^{2}= \sqrt{c^{2} -9}[/tex]

Hence, the expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]

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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 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 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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Answer:

1777777777888876332222233

In other words, 12 goes in the top box and 13 goes in the bottom. The fraction slash sign is not part of either box.

===========================================================

Explanation:

Cosine is the ratio of adjacent over hypotenuse.

cos(angle) = adjacent/hypotenuse

For the reference angle X, the adjacent leg is 36 units long. It's the leg closest or touching angle X.

The hypotenuse is always the longest side. It is always opposite the 90 degree angle. The hypotenuse in this case is 39 units.

Therefore,

cos(X) = 36/39 = (12*3)/(13*3) = 12/13

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 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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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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The dimension that would give the maximum area is 20.8569

Let the shorter side be = x

Perimeter of the semi-circle is πx

Twice the Length of the longer side

[tex][70-(\pi )x -x][/tex]

Length = [tex][70-(1+\pi )x]/2[/tex]

Total area =

area of rectangle + area of the semi-circle.

Total area =

[tex]x[[70-(1+\pi )x]/2] + [(\pi )(x/2)^2]/2[/tex]

When we square it we would have

[tex]70x +[(\pi /4)-(1+\pi)]x^2[/tex]

This gives

[tex]70x - [3.3562]x^2[/tex]

From here we divide by 2

[tex]35x - 1.6781x^2[/tex]

The maximum side would be at

[tex]x = 35/2*1.6781[/tex]

This gives us 20.8569

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

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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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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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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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

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.

The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³

An equation is an expression that shows the relationship between two or more numbers and variables.

The volume of a cylinder with radius (r) and height (h) is:

Volume = πr²h

Given that h = 10 ft, r = 16/2 = 8 ft

The volume = π * 8² * 10 = 2010.62 ft³

The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³

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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]

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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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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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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The true statements about terms in geometry that are undefinable are;

Three undefinable terms in geometry are;

1) Point

2) Line

3) Plane

The above terms do not have a formal definition but they can be described based on their properties.

The other geometric terms are defined based on the above undefinable terms.

A point can be described as a location in space that is dimensionless and can be specified on the coordinate plane as an ordered pairs (x, y).

True statements about a point are therefore;

A line is infinitely long, that has no beginning or end. It has one dimension, with no thickness or height.

Given that a point is dimensionless, a line can be considered a set of points.

A true statement is therefore;

A plane is a two dimensional geometric figure that have infinite length and width.

An infinite set of lines that forms a two dimensional figure can be used to describe a plane.

The true statement with regards to a plane is therefore;

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Answer:

Step-by-step explanation:

hello :

a(x) = 3x - 12 and b(x) = x-9, so

a[b(x)]=a(x-9) =3(x-9)-12

a[b(x)]=3x-9-12

a[b(x)]=3x+21

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 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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