Write the equation and
solve for x.
(4n+22) (8n-10)

Answer:

This can't be simplified anymore...its 8x^2

Step-by-step explanation:

This is a 2nd degree polynomial (also called a quadratic polynomial) since the highest power (or exponent) of a variable that occurs in the polynomial is the power 2.

Hope this helps:)

Answer:

the answer is (8x^2)

Step-by-step explanation:

Answer:

so the(g) semicircle should go up top with

The solution in interval notation b ∈ (-∞ , -7)

-5 times b is no less than 35.

5b  ≥ 35

n order to isolate the variable in this linear inequality, we need to get rid of the coefficient that multiplies it.

-\(\frac{5b}{5} = \frac{-35}{5}\)

This can be accomplished if both sides are divided by .

Notice that the inequality sign has changed.

b ≤ -7

We need to reduce this fraction to the lowest terms.

This can be done by dividing out those factors that appear both in the numerator and in the denominator.

In our example, this is the common factor:5

b ∈ (-∞ , -7)

We have already found the solution to the inequality, but since we want to represent the solution in interval notation, we have changed it.

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Answer:

i think it is the 2nd one and the 5th one.

Step-by-step explanation:

hope this helped!

p.s it would be cool if you gave me brainliest

The t procedure should not be used when there is a large outlier in the data or when the distribution shows moderate skewness. In these circumstances, the t procedure may not provide accurate results.

The t procedure assumes that the data is normally distributed. However, it can still be used under certain deviations from normality. The t procedure is robust to small departures from normality, so in the case of moderate skewness (option A), it can still provide reasonably accurate results. Skewness refers to the asymmetry of the distribution, and if it is only moderately skewed, the t procedure can be used.

However, there are situations where the t procedure should not be used. One such circumstance is when there is a large outlier in the data (option C). An outlier is an extreme value that differs significantly from the other observations. Large outliers can have a significant impact on the results of the t procedure, as it is sensitive to extreme values. In such cases, using the t procedure may lead to biased estimates or incorrect inferences.

Additionally, the sample standard deviation being large (option D) does not necessarily make the t procedure inappropriate. The t procedure is designed to handle variability in the data, including cases with larger standard deviations. As long as the other assumptions of the t procedure, such as normality and independence, are met, it can still be used effectively.

In summary, the t procedure should not be used when there is a large outlier in the data or when the distribution shows significant skewness. These situations can undermine the assumptions of the t procedure and may lead to inaccurate results.

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Answer: Circle

Step-by-step explanation:

When a\(_{n}\) = \(2^{n}\)+ n,  a₀ = 1,  a₁ = 3,  a₂ = 6, and a₃ = 11  

When a\(_{n}\) = n^(n+1)!,  a₀ = 0,  a₁ = 2,  a₂ = 2⁶, and a₃ = 3²⁴  

When a\(_{n}\) =  [n/2], a₀ = 0,  a₁ = 1/2,  a₂ = 1, and a₃ = 3/2      

When a\(_{n}\) = [n/2] + [n/2], a₀ = 0,  a₁ = 1,  a₂ = 2, and a₃ = 3/2  

Number sequence

A number sequence is a progression or a list of numbers that are directed by a pattern or rule.

Here,

a₀, a₁, a₂, and a₃ are terms of a sequence

from option a, a\(_{n}\) = \(2^{n}\)+ n

⇒ a₀  = 2⁰+ 0 = 1+0 = 1

⇒ a₁  = 2¹+ 1 = 2+1 = 3

⇒ a₂, = 2²+ 2 = 4+2 = 6

⇒ a₃  = 2³+ 3 = 8 +3 = 11  

from option b, a\(_{n}\) = n^(n+1)!

⇒ a₀  = 0^(0+1)! = 0

⇒ a₁  = 1^(1+1)! = 2² = 2

⇒ a₂, = 2^(2+1)! = 2^(3)! = 2⁶          [ ∵ 3! = 6 ]

⇒ a₃  = 3^(3+1)! = 3^(4)! = 3²⁴         [ ∵ 4! = 24 ]

from option c, a\(_{n}\) =  [n/2]          

⇒ a₀  = [0/2] = 0

⇒ a₁  = [1/2] = 1/2

⇒ a₂, = [2/2] = 1

⇒ a₃  = [3/2] =  3/2  

from option d, a\(_{n}\) = [n/2] + [n/2]

⇒ a₀  =  [0/2] + [0/2] = 0

⇒ a₁  =  [1/2] + [1/2] = 1/2 + 1/2 = 1

⇒ a₂, = [2/2] + [2/2] = 1 + 1 = 2

⇒ a₃  =  [3/2] + [3/2] = 6/4 = 3/2

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The Complete Question is -

What are the terms a₀, a₁, a₂, and a₃ of the sequence {a\(_{n}\)}, where a\(_{n}\) is where a\(_{n}\) equals

a. \(2^{n}\) + n                    b. n^(n+1)!

c. [n/2]                      d. [n/2] + [n/2]

Answer:

-3/3

Step-by-step explanation: rise over run the red line the rise goes up by three and the blue the run goes over by three but the line in going like this \ so the slope is negative

Answer:

V=168ft³

Step-by-step explanation:

V=

w-6

h-4

l-7  

V = 168ft³

To find the vertical asymptotes of a rational function, we need to set the denominator equal to zero and solve for x.

To find the vertical asymptotes of a rational function, you need to determine the values of x that make the denominator of the function equal to zero.

Let's check the limit of our example function as x approaches each of the vertical asymptotes:

As x approaches 1 from the left side: f(x) approaches negative infinity

As x approaches 1 from the right side: f(x) approaches positive infinity

Therefore, x = 1 is a valid vertical asymptote.

As x approaches 3 from the left side: f(x) approaches positive infinity

As x approaches 3 from the right side: f(x) approaches negative infinity

Therefore, x = 3 is also a valid vertical asymptote.

The values of x that make the denominator zero are the vertical asymptotes of the function.

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Answer:

herlojffhgdxvjdxhyeb olis the tim view on the euros 2020 final

Equation -1 = 0 doesn't have solution, no value of r that satisfies all coefficients. Equation r f(x, y) + y f(x, y) = 3 not satisfied for any value of r. Equation r f(x, y) + y f(x, y) = 3 doesn't hold no value of r that satisfies it.

In this problem, we are given a function f: R² → R defined as follows:

f(x, y) = 5x² if (x, y) = (0, 0)

f(x, y) = -x² + 7y² if (x, y) ≠ (0, 0)

We need to determine if the function f has a limit at (0, 0) and analyze its continuity.

(a) To determine if the function f has a limit at (0, 0), we need to compute the limit of f(x, y) as (x, y) approaches (0, 0) along different paths.

Along the x-axis: Letting y = 0, we have f(x, 0) = 5x². Taking the limit as x approaches 0, we get lim(x→0) f(x, 0) = lim(x→0) 5x² = 0.

Along the y-axis: Letting x = 0, we have f(0, y) = -x² + 7y² = 7y². Taking the limit as y approaches 0, we get lim(y→0) f(0, y) = lim(y→0) 7y² = 0.

Since the limit of f(x, y) approaches 0 along both the x-axis and the y-axis, we can conclude that the function f has a limit at (0, 0).

(b) To determine the set of points for which f is continuous, we need to consider the function's definition at (0, 0) and its definition for all other points.

At (0, 0), the function is defined as f(0, 0) = 5x².

For all other points (x, y) ≠ (0, 0), the function is defined as f(x, y) = -x² + 7y².

Therefore, the set of points for which f is continuous is R², except for the point (0, 0) where f has a removable discontinuity.

(c) To show that r f(x, y) + y f(x, y) = 3, we substitute the given definitions of f(x, y) into the equation:

r f(x, y) + y f(x, y) = r(5x²) + y(-x² + 7y²)

= 5rx² - xy² + 7y³

Now, we need to find the value of r such that the expression equals 3. Setting the expression equal to 3, we have:

5rx² - xy² + 7y³ = 3

To find r, we can equate the coefficients of like terms on both sides of the equation. Comparing the coefficients, we get:

5r = 0 (coefficient of x² term)

-1 = 0 (coefficient of xy² term)

7 = 0 (coefficient of y³ term)

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\(~~~~~~ \textit{Simple Interest Earned}\\\\I = Prt\qquad\begin{cases}I=\textit{interest earned}\dotfill &\$2700\\P=\textit{original amount deposited}\dotfill & \$5000\\r=rate\to r\%\to \frac{r}{100}\dotfill &0.0r\\t=years\dotfill &9\end{cases}\\\\\\2700=(5000)(0.0r)(9)\implies 2700=45000(0.0r)\implies \cfrac{2700}{45000}=0.0r\\\\\\\cfrac{3}{50} = 0.0r\implies \cfrac{3}{50} = \cfrac{r}{100}\implies \cfrac{3\cdot 100}{50}=r\implies 6 = \stackrel{\%}{r}\)

An algebraic expression like (3+5) x (4-1) is a combination of numbers and at least one operation.

Answer: 237.160283353

Step-by-step explanation:

Answer:

the answer is 237.1602834

source: Calculator

I will win about $ 59.

This is a question of permutations and combinations.

We know that,

Probability of getting two non face cards = \(\frac{^{40}C_2}{{^{52}C_2}}\) = 10/17

Probability of not getting two non face cards = 1 - (10/17) = 7/17

Hence, we can write,

Money I will get = (7/17)*50*20 =  $ 411.764

Money I'll have to pay = (10/17)*30*20 = $ 352.94

We know that,

Money left with me = Money I will get - Money I'll have to pay

Hence, we can write,

Money left with me =   $ 411.764 - $ 352.94

Money left with me =  58.824 ≈ $ 59

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Answer:

11. 15°

I haven't learned how to find sides with only one side given yet I'm so sorry!!

To prove that for quadratic Bezier curves, the slope of the line segment p0q equals the slope of the tangent line to the curve at p0, and that the slope of the line segment qp1 equals the slope of the tangent line to the curve at p1, we'll use the properties of Bezier curves and their derivatives.

A quadratic Bezier curve is defined by three control points: p0, p1, and p2. The curve is parameterized by a variable t, where t ranges from 0 to 1. The equation for the quadratic Bezier curve is:

B(t) = (1 - t)^2 * p0 + 2 * (1 - t) * t * p1 + t^2 * p2

Let's calculate the derivative of the Bezier curve with respect to t, denoted as B'(t):

B'(t) = d/dt [(1 - t)^2 * p0 + 2 * (1 - t) * t * p1 + t^2 * p2]

= -2 * (1 - t) * p0 + 2 * (1 - 2t) * p1 + 2 * t * p2

= -2 * (1 - t) * p0 + 2 * (1 - 2t) * p1 + 2 * t * p2

Now, let's find the slope of the line segment p0q, where q is any point on the curve. We'll substitute t = 0 into the equation of the Bezier curve:

q = B(0) = (1 - 0)^2 * p0 + 2 * (1 - 0) * 0 * p1 + 0^2 * p2

= p0

Therefore, q = p0.

The slope of the line segment p0q is given by the difference in y-coordinates divided by the difference in x-coordinates:

slope_p0q = (q.y - p0.y) / (q.x - p0.x)

Since q = p0, the numerator becomes 0, and the slope of the line segment p0q becomes 0.

Now, let's find the slope of the tangent line to the curve at p0. We'll substitute t = 0 into the equation for the derivative of the Bezier curve:

B'(0) = -2 * (1 - 0) * p0 + 2 * (1 - 2 * 0) * p1 + 2 * 0 * p2

= -2 * p0 + 2 * p1

The slope of the tangent line to the curve at p0 is given by the y-component divided by the x-component of the derivative at t = 0:

slope_tangent_p0 = (B'(0)).y / (B'(0)).x

= (-2 * p0 + 2 * p1).y / (-2 * p0 + 2 * p1).x

Since we have q = p0, we can simplify the expression:

slope_tangent_p0 = (-2 * q + 2 * p1).y / (-2 * q + 2 * p1).x

Now, let's find the slope of the line segment qp1. We'll substitute t = 1 into the equation of the Bezier curve:

q = B(1) = (1 - 1)^2 * p0 + 2 * (1 - 1) * 1 * p1 + 1^2 * p2

= p2

Therefore, q = p2.

The slope of the line segment qp1 is given by the difference in y-coordinates divided by the difference in x-coordinates:

slope_qp1 = (p2.y - q.y) / (p2.x - q.x)

Since q = p2, the numerator becomes 0, and the slope of the line segment qp1 becomes 0.

Now, let's find the slope of the tangent line to the curve at p1. We'll substitute t = 1 into the equation for the derivative of the Bezier curve:

B'(1) = -2 * (1 - 1) * p0 + 2 * (1 - 2 * 1) * p1 + 2 * 1 * p2

= 2 * p2 - 2 * p1

The slope of the tangent line to the curve at p1 is given by the y-component divided by the x-component of the derivative at t = 1:

slope_tangent_p1 = (B'(1)).y / (B'(1)).x

= (2 * p2 - 2 * p1).y / (2 * p2 - 2 * p1).x

Since we have q = p2, we can simplify the expression:

slope_tangent_p1 = (2 * q - 2 * p1).y / (2 * q - 2 * p1).x

As we can see, both slope_p0q and slope_tangent_p0 are equal to 0, and both slope_qp1 and slope_tangent_p1 are equal to 0.

Therefore, we have proved that for quadratic Bezier curves, the slope of the line segment p0q equals the slope of the tangent line to the curve at p0, and the slope of the line segment qp1 equals the slope of the tangent line to the curve at p1.

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Answer:

h = 440.94 feet

Step-by-step explanation:

It is given that,

An architect is standing 370 feet from the base of a building, x = 370 feet

The angle of elevation is 50°.

We need to find the approximate height of the building. let it is h. It can be calculated using trigonometry as follows :

\(\tan\theta=\dfrac{P}{B}\\\\\tan\theta=\dfrac{h}{x}\\\\h=x\tan\theta\\\\h=370\times \tan50\\\\h=440.94\ \text{feet}\)

So, the approximate height of the building is 440.94 feet

We have to use the equation

Where r = 12 mph, d = 20 miles, let's find t.

She'll take 5/3 hours to get there which is equivalent to 1 2/3 hours.

Let's transform 2/3 hours to minutes to find the exact time she has to leave.

We know that 1 hour is 60 minutes.

So, she will take exactly 1 hour and 40 minutes.

The balance in your account after 10 years will be $2,473.08.

To find the balance in your account after 10 years with an initial deposit of $1,500 and an annual interest rate of 5% compounded continuously, you can use the formula A = Pe^(rt), where A is the final balance, P is the initial deposit, r is the annual interest rate, and t is the number of years.

In this case, P = $1,500, r = 5%, and t = 10. Plugging these values into the formula, we get:

A = $1,500 * e^(0.05 * 10)
A = $1,500 * e^(0.5)
A = $1,500 * 1.64872
A = $2,473.08

the balance in your account after 10 years will be $2,473.08.

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20 units is the measure of the length of IK.

A line is defined as the distance between two points. Given the following parameters

IK =5x

IJ = 4

JK =4x

If the point J is on IK, hence;

IK = IJ + JK

Substitute the given parameters to have:

5x = 4 + 4x

5x - 4x = 4

x = 4

Determine the length of IK

IK = 5x = 5(4)

IK = 20

Hence the measure of the length of IK is 20 units
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Complete question

Point J is on line segment IK.Given IK =5x, IJ = 4 and JK =4x, determine the numerical length of IK

Step-by-step explanation:

Eli typed faster on his second attempt, 120 words per minute on his first attempt and 150 words per minute on his second attempt.

A fraction represents a part of a whole.

We need to calculate his typing speed for each attempt.

On his first attempt, Eli typed 180 words in \(1\frac{1}{2}\) hours, which is the same as typing 120 words in 1 hour.

Therefore, his typing speed on the first attempt was 120 words per minute.

On his second attempt, Eli typed 180 words in \(1\frac{1}{5}\) hour, which is the same as typing 150 words in 1 hour.

Therefore, his typing speed on the second attempt was 150 words per minute.

Comparing the two speeds, we see that Eli was faster on his second attempt, typing 150 words per minute compared to 120 words per minute on his first attempt.

Therefore, Eli typed faster on his second attempt, 120 words per minute on his first attempt and 150 words per minute on his second attempt.

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Answer:-y,x

Step-by-step explanation: It is rotateing into the negative part of a graph if you think about it.

Answer: The answer is D.

Step-by-step explanation: i did the test

Answer:

Step-by-step explanation:

What shape? A right cone?

√(16^2+25^2) =√881 ≈ 29.7cm

The length of the slant height is 29.68 centimeters.

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides, according to Pythagoras's Theorem.

Given:

The radius, r, of the base is 16 cm.

The height, h, is 25 cm.

To find the length of slant height:

Using the Pythagorean theorem,

l = √(25)² + (16)²

l = 29.68 centimeters.

Therefore, slant height is 29.68 centimeters.

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The homogeneous equation and a particular solution of the given nonhomogeneous equation is \(& y=c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}\).

The given nonhomogeneous equation is y" - 4y = 2

We are given \($y_1=e^{-2 x}$\)

Let \($y_2=u y_1$\)

Non-homogeneous differential equations are simply differential equations that do not satisfy the conditions for homogeneous equations. In the past, we’ve learned that homogeneous equations are equations that have zero on the right-hand side of the equation.

The two most common methods when finding the particular solution of a non-homogeneous differential equation are:

\($$\begin{aligned}& y_2=u e^{-2 x} \\& y_2^{\prime}=u^{\prime} e^{-2 x}-2 u e^{-2 x} \\& y_2^{\prime \prime}=u^{\prime \prime} e^{-2 x}-2 u^{\prime} e^{-2 x}-2 u^{\prime} e^{-2 x}+4 u e^{-2 x} \\& y_2^{\prime \prime}=u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}+4 u e^{-2 x}\end{aligned}$$\)

Substitute in \($y^{\prime \prime}-4 y=0$\), we get

\($$\begin{aligned}& u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}+4 u e^{-2 x}-4 u e^{-2 x}=0 \\& u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}=0 \\& e^{-2 x}\left(u^{\prime \prime}-4 u^{\prime}\right)=0 \\& u^{\prime \prime}-4 u^{\prime}=0 \\& u^{\prime \prime}=4 u^{\prime}\end{aligned}$$\)

Substitute \($$u^{\prime}=v$ and $u^{\prime \prime}=v^{\prime}$\), we get

\($$\begin{aligned}& v^{\prime}=4 v \\& \frac{d v}{d x}=4 v \\& \frac{1}{v} d v=4 d x\end{aligned}$$\)

Integrate both sides, we get

\($u=\frac{1}{4} C_1 e^{4 x}+C_2$$\)

put in \($y_2=u y_1$\)

\($$y_2=\left(\frac{1}{4} C_1 e^{4 x}+C_2\right) e^{-2 x}$$\)

Choose \($$C_1=4$ and $C_2=0$\), we get

\($$y_2=e^{2 x}$$\)

\($$\begin{aligned}& y_c=c_1 y_1+c_2 y_2 \\& y_c=c_1 e^{-2 x}+c_2 e^{2 x}\end{aligned}$$\)

Particular Solution

\($$\begin{aligned}& y_y=A \\& y_y{ }^{\prime}=0 \\& y_y{ }^{\prime \prime}=0\end{aligned}$$\)

Substitute in \($y^{\prime \prime}-4 y=2$\), we get

\($$\begin{aligned}& 0-4 A=2 \\& -4 A=2 \\& A=-\frac{1}{2} \\& \therefore y_y=-\frac{1}{2}\end{aligned}$$\)

General Solution

\($$\begin{aligned}& y=y_6+y_y \\& y=c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}\end{aligned}$$\)

Therefore, the second equation is \(& y_{2} =c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}\).

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Step 1: Suppose, for the sake of contradiction, that there are integers A and B such that A2 = 3B2 + 2015. Let N = A2. Then, N ≡ 1 (mod 3).

Step 2: By the Legendre symbol, since (2015/5) = (5/2015) = -1 and (2015/67) = (67/2015) = -1, we know that there is no integer k such that k2 ≡ 2015 (mod 335).

Step 3: Let's consider A2 = 3B2 + 2015 (mod 335). This can be written as A2 ≡ 195 (mod 335), which can be further simplified to N ≡ 1 (mod 5) and N ≡ 3 (mod 67).

Step 4: However, since (2015/5) = -1, it follows that N ≡ 4 (mod 5) is a contradiction.

Therefore, there are no integers A, B such that A2 = 3B2 + 2015.

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Answer and explanation:

When you use positive interger to represent income, it shows that there has been an addition instead of a subtraction in money value to the business. In other words money is coming in instead of going out. Example when there is profit on goods sold say $10, it is indicated with a positive integer 10 to indicate addition in terms of operating income

When you use a negative value to represent expense, it indicates there has been a subtraction instead of an addition in money value. In other words money is going out of the business and not coming in. Example when there is a transportation expense to convey goods to the buyer, it is recorded as expenses subtracted from say revenue sales of the business. expenses are usually indicated in brackets in accounting to show negative value, although not a requirement since they are already named and recognized as expenses and subtracted regardless