Let X1 and X2 be two stochastically independent random variables so that the variances of X1 and X2 are (\sigma1)2 = k and (\sigma2)2 = 2, respectively. Given that the variance of Y = 3X2 - X1 is 25, find k.

Answer: w < 2,200

Step-by-step explanation:

The value of p is 20/19, 19/16.

What is the distance formula?

Using their coordinates, an algebraic expression provides the distances between two points (see coordinate system). The distance formula is a formula used to determine how far apart two places are from one another. The dimensions of these points are unlimited.

Here, we have

Given: The distance between the points (1, 2p) and (1- p, 1) is 11-9p.

We have to find the value of p.

AB = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

11 - 9p = [tex]\sqrt{(1-p-1)^2 +(1-2p)^2}[/tex]

(11 - 9p)² = 5p² - 4p + 1

121 + 81p² - 198p =  5p² - 4p + 1

121 + 81p² - 198p -  5p² + 4p - 1 = 0

120 + 76p² - 194p = 0

76p² - 194p + 120 = 0

38p² - 97p + 60 = 0

p = 20/19, 19/16

Hence, the value of p is 20/19, 19/16.

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The value of p is 20/19, 19/16.

What is the distance formula?

Using their coordinates, an algebraic expression provides the distances between two points (see coordinate system). The distance formula is a formula used to determine how far apart two places are from one another. The dimensions of these points are unlimited.

Here, we have

Given: The distance between the points (1, 2p) and (1- p, 1) is 11-9p.

We have to find the value of p.

AB = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

11 - 9p = [tex]\sqrt{(1-p-1)^2 +(1-2p)^2}[/tex]

(11 - 9p)² = 5p² - 4p + 1

121 + 81p² - 198p =  5p² - 4p + 1

121 + 81p² - 198p -  5p² + 4p - 1 = 0

120 + 76p² - 194p = 0

76p² - 194p + 120 = 0

38p² - 97p + 60 = 0

p = 20/19, 19/16

Hence, the value of p is 20/19, 19/16.

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A polynomial that satisfies the given conditions is f(x) = a(x + 2)^2(x - 2)^2x^3, where a is a constant.

To find the polynomial f(x) that meets the given requirements, we can start by noting that since -2 and 2 are zeros of multiplicity 2, the factors (x + 2)^2 and (x - 2)^2 must be included in the polynomial. Additionally, since 0 is a zero of multiplicity 3, the factor x^3 must also be included.

So far, we have the polynomial in the form f(x) = a(x + 2)^2(x - 2)^2x^3, where a is a constant that we need to determine.

To find the value of a, we can use the fact that f(-1) = 45. Plugging in x = -1 into the polynomial, we get:

f(-1) = a(-1 + 2)^2(-1 - 2)^2(-1)^3

= a(1)^2(-3)^2(-1)

= 9a

Setting 9a equal to 45, we can solve for a:

9a = 45

a = 5

So the polynomial f(x) that satisfies the given conditions is:

f(x) = 5(x + 2)^2(x - 2)^2x^3.

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+) Case 1: a = 0

=> f(x) = 0×x²+c = c

=> for all values of x, f(x) always = c (does not satisfy the requirement)

+) Case 2: a≠0

=> f(x) = ax²+c

=> for every non-zero a, f(x) has only one solution x

Ans: a≠0, c ∈ R

P/s: c can be any value (in case you don't know the symbols above)

Ok done. Thank to me >:333

Your  annual salary of $12.500 increases by 3% each year after 5 years would be $14,456.47.

What is function?

In mathematics, a function is a relation between a set of inputs and a set of possible outputs with the property that each input is related to exactly one output. A function is typically denoted by a symbol, such as f(x), where "f" is the name of the function and "x" is the input variable. The output of the function is obtained by applying a rule or formula to the input variable.

The function that represents your salary after x years can be written as:

y = 12500(1 + 0.03)ˣ

where y is your salary in dollars after x years, 12500 is your starting salary in dollars, and 0.03 is the annual increase rate as a decimal (3% = 0.03).

To calculate your salary after, say, 5 years, you would substitute x = 5 into the function:

y = 12500(1 + 0.03)

y = 14,456.47

Therefore, Your  annual salary of $12.500 increases by 3% each year after 5 years would be $14,456.47.

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According to the frequency distribution for the life-length statistics, 

a) assuming the manufacturer's claim is accurate, 84% of grade A batteries will survive longer than 50 months.

b) About 8.2% of the manufacturer's batteries will last less than 40 months, assuming their claim is true.

a) Assuming the manufacturer's claim is true, the distribution of the battery life length will be normal with a mean of 60 months and a standard deviation of 10 months.

To find the percentage of batteries that will last more than 50 months, we need to find the area under the normal curve to the right of x = 50.

Using a standard normal distribution table or a calculator, we can find that the area to the right of z = (50-60)/10 = -1 is approximately 0.8413. The manufacturer's grade A batteries will therefore last beyond 50 months for about 84.13% of them.

b) Again assuming the manufacturer's claim is true, to find the percentage of batteries that will last less than 40 months, we need to find the area under the left of the x = 40 normal curve.

Using the same method as in part a), we find that the area to the left of z = (40-60)/10 = -2 is approximately 0.0228.

Therefore, approximately 2.28% of the manufacturer's batteries will last less than 40 months.

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The maximum of z if: x+y+z =5, xy+yz +xz  is 3.

We may determine z from the first equation by resolving it in terms of x and y:

z = 5 - x - y

With the second equation as a substitute

5 - x - y + xy + y(5 - x - y) + x = 3

2xy - 5y - 5x + 25 = 0

Solving for y

y=[5 √(25 - 8x)] / 4

So,

x = y = 1

Adding back into the initial equation

z = 5 - x - y = 3

Therefore the maximum value of z is 3 which occurs when x = y = 1.

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Mr Tan is 69 years old now.

Let's denote Mr. Tan's age as T and Peiling's age as P.

We are given two pieces of information:
5 years ago, Mr Tan was 4 times as old as Peiling.
Peiling is 48 years younger than Mr Tan now.
Now let's translate these into equations:
T - 5 = 4 * (P - 5)
P = T - 48
Next, we'll solve for P in equation 1 and then substitute it into equation 2:
T - 5 = 4 * (P - 5)
T - 5 = 4P - 20
4P = T + 15
Now, substitute P from equation 2 into this equation:
4 * (T - 48) = T + 15
4T - 192 = T + 15
3T = 207
T = 69.

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The main answer is that by plugging y into the differential equation, we get:

y' + 2y = (2/3)eˣ + e⁻²ˣ + 2(2/3)eˣ + 2e⁻²ˣ

Simplifying this expression, we get:

y' + 2y = (8/3)eˣ + (3/2)e⁻²ˣ

And since this is equal to 2eˣ, we can see that y is a solution of the differential equation.

The explanation is that in order to show that y is a solution of the differential equation, we need to plug y into the equation and see if it satisfies the equation.

In this case, we get an expression that simplifies to 2eˣ, which is the same as the right-hand side of the equation. Therefore, we can conclude that y is indeed a solution of the differential equation. This method is commonly used to verify solutions of differential equations and is a useful tool for solving more complex problems.

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The answer is False. In a finite field, for each non-zero number x, there exists a multiplicative inverse y such that x*y = 1, not 0. The only number that would satisfy x*y = 0 in a finite field is when either x or y is 0.

True. In a finite field, every non-zero element has a multiplicative inverse, meaning that there exists a number y such that x*y = 1. Therefore, if we multiply both sides by 0, we get x*(y*0) = 0, which simplifies to x*0 = 0. Therefore, for each number x in a finite field, there is a number y such that x*y = 0.
Multiplying an even number equals dividing by its difference and vice versa. For example, dividing by 4/5 (or 0.8) will give the same result as dividing by 5/4 (or 1.25). That is, multiplying a number by its inverse gives the same number (because the product and difference of a number are 1.

The term reciprocal is used to describe two numbers whose product is 1, at least in the third edition of the Encyclopedia Britannica (1797); In his 1570 translation of Euclid's Elements, he mutually defined inversely proportional geometric quantities.

In the multiplicative inverse, the required product is usually removed and then understood by default (as opposed to the additive inverse). Different variables can mean different numbers and numbers. In these cases, it will appear as

ab ≠ ba; then "reverse" usually means that an element is both left and right reversed.

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The general solution to the differential equation 6y'' + 6y = 2tan(6) - 1/2e^{3t} is y(t) = c1*cos(t) + c2*sin(t) + (1/6)tan(6) - (1/36)e^{3t}.

To find this solution, first, solve the homogeneous equation 6y'' + 6y = 0. The characteristic equation is 6r^2 + 6 = 0. Solving for r gives r = ±i.

The homogeneous solution is y_h(t) = c1*cos(t) + c2*sin(t), where c1 and c2 are constants. Next, find a particular solution y_p(t) for the non-homogeneous equation by using an ansatz. For the tan(6) term, use A*tan(6), and for the e^{3t} term, use B*e^{3t}.

After substituting the ansatz into the original equation and simplifying, we find that A = 1/6 and B = -1/36. Thus, y_p(t) = (1/6)tan(6) - (1/36)e^{3t}. Finally, combine the homogeneous and particular solutions to get the general solution: y(t) = c1*cos(t) + c2*sin(t) + (1/6)tan(6) - (1/36)e^{3t}.

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We can say with 90% confidence that the true mean level of polyunsaturated fatty acid in this brand of diet margarine is between 16.95 and 17.23. The answer is 90%.

Using the t-distribution with 5 degrees of freedom (n-1), we can calculate the t-value for a 90% confidence interval. We use a one-tailed test because we want to find the confidence interval for values greater than 16.65:

t-value = t(0.90,5) = 1.476

Now we can calculate the margin of error (E) for a 90% confidence interval:

E = t-value * (s / √n) = 1.476 * (0.31 / √6) = 0.28

Finally, we can calculate the confidence interval:

16.95 + E = 16.95 + 0.28 = 17.23

Therefore, we can say with 90% confidence that the true mean level of polyunsaturated fatty acid in this brand of diet margarine is between 16.95 and 17.23. Since the range of values between 16.65 and 17.32 falls within this confidence interval, we can also say that we are 90% confident that the true mean level of polyunsaturated fatty acid falls within this range.

So, the answer is 90%.

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The indefinite integral of sin^4(5θ) dθ is (1/4)[θ - sin(10θ)/5 + (θ/2) + (1/40)sin(20θ)] + C.

An indefinite integral is the reverse operation of differentiation. Given a function f(x), its indefinite integral is another function F(x) such that the derivative of F(x) with respect to x is equal to f(x), that is:

F'(x) = f(x)

The symbol used to denote the indefinite integral of a function f(x) is ∫ f(x) dx. The integral sign ∫ represents the process of integration, and dx indicates the variable of integration. The resulting function F(x) is also called the antiderivative or primitive of f(x), and it is only unique up to a constant of integration. Therefore, we write:

∫ f(x) dx = F(x) + C

where C is an arbitrary constant of integration. Note that the indefinite integral does not have upper and lower limits of integration, unlike the definite integral.

We can use the identity [tex]sin^2(x)[/tex] = (1/2)(1 - cos(2x)) to simplify the integrand:

[tex]sin^4[/tex](5θ) = ([tex]sin^2[/tex](5θ)[tex])^2[/tex]

= [(1/2)(1 - cos(10θ))[tex]]^2[/tex] (using [tex]sin^2[/tex](x) = (1/2)(1 - cos(2x)))

= (1/4)(1 - 2cos(10θ) +[tex]cos^2[/tex](10θ))

Expanding the square and integrating each term separately, we get:

∫ [tex]sin^4[/tex](5θ) dθ = (1/4)∫ (1 - 2cos(10θ) + [tex]cos^2[/tex](10θ)) dθ

= (1/4)[θ - sin(10θ)/5 + (1/2)∫ (1 + cos(20θ)) dθ] + C

= (1/4)[θ - sin(10θ)/5 + (θ/2) + (1/40)sin(20θ)] + C

Therefore, the indefinite integral of sin^4(5θ) dθ is (1/4)[θ - sin(10θ)/5 + (θ/2) + (1/40)sin(20θ)] + C.

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Answer:

60

Step-by-step explanation:

The arc cosine of 0.5 can be written as cos⁻¹(0.5). To find its value in degrees, we can use a calculator or reference table. Specifically, we have:

cos⁻¹(0.5) ≈ 60 degrees

The direction of the resultant vector is determined as -21.80⁰.

The value of angle between the two vectors is the direction of the resultant vector and it is calculated as follows;

tan θ = vy/vx

where;

( -4, 12), (6, 8)

vy = (8 - 12) = -4

vx = (6 + 4) = 10

tan θ = ( -4 ) / ( 10 )

tan θ = -0.4

The value of θ is calculated  by taking arc tan of the fraction,;

θ = tan ⁻¹ ( -0.4 )

θ =  -21.80⁰

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The f-statistic use to test the hypothesis.is 34.719264.

To calculate the F-statistic for testing the overall significance of the linear regression model,

we need to compare the regression sum of squares (SSR) to the residual sum of squares (SSE) and the degrees of freedom associated with each.

The formula for the F-statistic is:

F = (SSR / k) / (SSE / (n - k - 1))

where k is the number of predictor variables in the model, and n is the sample size.

Since the question does not provide the values of k and n, I will assume that k = 1 (simple linear regression) and use the information given in the previous question to find n.

From the previous question, we have:

SSE = 24.074572

MSE = SSE / (n - 2) = 2.674952

SSTO = SSR + SSE = 83.820408

R-squared = SSR / SSTO = 0.7136

We can use R-squared to find SSTO:

SSTO = SSR / R-squared = 83.820408 / 0.7136 = 117.539337

Then, we can use SSTO and MSE to find n:

SSTO / MSE = n - 2

117.539337 / 2.674952 = n - 2

n = 45

Now we can substitute the values of k, n, SSR, and SSE into the formula for the F-statistic:

F = (SSR / k) / (SSE / (n - k - 1))

F = ((117.539337 - 83.820408) / 1) / (24.074572 / (45 - 1 - 1))

F = 34.719264

Therefore, the F-statistic is 34.719264.

This value can be used to test the hypothesis that the slope coefficient is equal to zero, with a significance level determined by the degrees of freedom and the chosen alpha level.

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Answer:

Common ratio = r

r = [tex]\frac{a_2}{a_1} =\frac{3}{4}=0.75[/tex]

The probability that a student has a dog, given that they do not have a cat is 5/6.

The probability that a student has a dog given that they do not have a cat is :

P ( Has a dog | Does not have a cat ) = P ( Has a dog and does not have a cat) / P ( Does not have a cat )

Total number of students = 11 + 10 + 5 + 2 = 28

P ( Has a dog | Does not have a cat ):

= (10 / 28) / (12 / 28)

= 10 / 12

= 5 / 6

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Thus, the solution of the given system of equation is found as - (2,3).

The ordered pair that provides the solution to by both equations is the system's solution. We graph both equations using a single coordinate system in order to visually solve a system of linear equations. The intersection of the two lines is where the system's answer will be found.

The given system of equation are-

-3x - y = -9 ..eq 1

3x - y = 3  ..eq 2

Consider eq 1

-3x - y = -9

Put x = 0; -3(0) - y = -9 --> y = 9 ; (0,9)

Put y = 0; -3x - (0) = -9  ---> x = 3 ; (3,0)

Consider eq 1

3x - y = 3

Put x = 0; 3(0) - y = 3 --> y = -3 ; (0,-3)

Put y = 0; 3x - (0) = 3  ---> x = 1 ; (1,0)

Plot the obtained points on graph, the intersection points gives the solution of the system of equations.

Solution - (2, 3)

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The phasor form (in the rms sense) of the given expressions are:

a. 20∠(-180°) V

b. 6 x 10⁻⁶∠90° A

c. 3.6 x 10⁻⁶∠(-20°) A

a. The given expression is in the form of 20 sin (ωt - φ), where ω is the angular frequency and φ is the phase angle in degrees. To convert it to phasor form, we need to express it as a complex number in the form of Vrms∠θ, where Vrms is the root mean square (rms) value of the voltage and θ is the phase angle in radians. In this case, the rms value is 20 V and the phase angle is -180° (since it is given as -180° in the expression). The phasor form can be represented as 20∠(-180°) V.

b. The given expression is in the form of 6 x 10⁻⁶ cos(ωt), where ω is the angular frequency. To convert it to phasor form, we need to express it as a complex number in the form of Irms∠θ, where Irms is the rms value of the current and θ is the phase angle in radians. In this case, the rms value is 6 x 10^(-6) A and the phase angle is 90° (since it is cos(ωt)). The phasor form can be represented as 6 x 10⁻⁶∠90° A.

c. The given expression is in the form of 3.6 x 10⁻⁶ cos(ωt - φ), where ω is the angular frequency and φ is the phase angle in degrees. To convert it to phasor form, we need to express it as a complex number in the form of Irms∠θ, where Irms is the rms value of the current and θ is the phase angle in radians. In this case, the rms value is 3.6 x 10⁻⁶ A and the phase angle is -20° (since it is given as -20° in the expression). The phasor form can be represented as 3.6 x 10⁻⁶∠(-20°) A.

THEREFORE, the phasor form (in the rms sense) of the given expressions are:

a. 20∠(-180°) V

b. 6 x 10⁻⁶∠90° A

c. 3.6 x 10⁻⁶∠(-20°) A.

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The given equation for the domain is C=5.5b, where b represents the number of dollar bills produced and C represents the total cost of producing those dollar bills.

We are told that the mint produces at least 420 dollar bills but not more than 425 dollar bills. Therefore, the domain of the function C=5.5b for this situation is the set of values of b that satisfy this condition.

In interval notation, we can represent this domain as follows:

Domain: 420 ≤ b ≤ 425

Therefore, the domain of the function C=5.5b for this situation is 420 ≤ b ≤ 425.

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The Comparison Test tells us that the original series[tex]Σ(5^k / (3^k + 4^k))[/tex]from k=1 to infinity also diverges.

Based on your question, you want to determine if the series [tex]Σ(5^k / (3^k + 4^k))[/tex] from k=1 to infinity converges or diverges. To analyze this series, we can apply the Comparison Test.

Consider the series [tex]Σ(5^k / 4^k)[/tex]from k=1 to infinity. This simplifies to [tex]Σ((5/4)^k)[/tex], which is a geometric series with a common ratio of 5/4. Since the common ratio is greater than 1, this series diverges.

Now, notice that[tex]5^k / (3^k + 4^k) ≤ 5^k / 4^k[/tex] for all k≥1. Since the series [tex]Σ(5^k / 4^k)[/tex]diverges, and the given series is term-wise smaller, the Comparison Test tells us that the original series [tex]Σ(5^k / (3^k + 4^k))[/tex] from k=1 to infinity also diverges.

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The diameter of the tire based on the amplitude of the sinusoidal graph of the height of the tire valve, y inches above or below the center of the tire is 20 inches.

A sinusoidal function is a periodic function that is based on the sine or cosine function.

The shape of the graph is the shape of a sinusoidal function graph.

The y-coordinates of the peak and the through are; y = 10 and y = -10

The shape of the tire is a circle

The amplitude, A, of the sinusoidal function, which has a magnitude equivalent to the length of the radius of the tire is therefore;

A = (10 - (-10))/2 = 10

The radius of the tire, r = A = 10 inches

The diameter of the tire = 2 × The radius of the tire

The diameter of the tire = 2 × 10 inches = 20 inches

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The measurement of angle A is approximately 38.8 degrees and the measurement of angle B is approximately 51.2 degrees.

Triangles and the connections between their sides and angles are studied in the branch of mathematics known as trigonometry. Trigonometric functions like sine, cosine, and tangent are used to solve problems involving right triangles and other geometric shapes in a variety of disciplines, including science, engineering, and physics.

We can use trigonometry to solve for the angle A.

First, we can find the length of the hypotenuse AB using the Pythagorean theorem:

AB² = BC² + CA²

AB² = 19² + 22²

AB² = 905

AB = √(905)

AB = 30.1

Next, we can use the sine function to find the measure of angle A:

sin(A) = BC / AB

sin(A) = 19 / 30.1

A = sin⁻¹(19 / 30.1)

A = 38.8

Finally, we can use the fact that the sum of the angles in a triangle is 180 degrees to find the measure of angle B:

B = 90 - x

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The vertex form of the quadratic function y = x² - 6x - 7 is y = (x - 3)² - 16

Given the quadratic function in the question:

y = x² - 6x - 7

The vertex form of a quadratic function is expressed as:

y = a(x - h)² + k

Where (h, k) is the vertex of the parabola and "a" is a coefficient that determines the shape of the parabola.

To write y = x² - 6x - 7 in vertex form, we need to complete the square.

We can do this by adding and subtracting the square of half the coefficient of x:

y = x² - 6x - 7

y = (x² - 6x + 9) - 9 - 7 (adding and subtracting 9)

y = (x - 3)² - 16

Hence, the vertex form is:

y = (x - 3)² - 16

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Using Leibniz's rule, the final answer is xd/dx ∫ f(t) dt = x f(x) + ∫ f(t) dt + x f'(x)

Using Leibniz's rule, we have:

x d/dx ∫ f(t) dt = x f(x) + ∫ x d/dx f(t) dt

The first term x f(x) comes from differentiating the upper limit of integration with respect to x, while the second term involves differentiating under the integral sign.

If we assume that f(x) is a differentiable function, then by the chain rule, we have:

d/dx f(x) = d/dx [f(t)] evaluated at t = x

Therefore, we can rewrite the second term as:

∫ x d/dx f(t) dt = ∫ x d/dt f(t) dt evaluated at t = x

= ∫ f(t) dt + x f'(x)

Substituting this into the original equation, we obtain:

xd/dx ∫ f(t) dt = x f(x) + ∫ f(t) dt + x f'(x)

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All angles in the triangle are 60° and all sides are approximately 7.348.

Using the law of cosines, we can find side BC:

BC² = AB² + AC² - 2AB(AC)cos(α)
BC² = AB² + 9 - 2AB(9)cos(60°)
BC² = AB² + 9 - 9AB
BC² = 9 - 9AB + AB²

We also know that angle B is 60° (since it is an equilateral triangle). Using the law of sines, we can find AB:

AB/sin(60°) = AC/sin(B)
AB/sqrt(3) = 9/sin(60°)
AB/sqrt(3) = 9/√3
AB = 9

Substituting AB = 9 into the equation for BC², we get:

BC² = 9 - 9(9) + 9²
BC² = 54
BC = sqrt(54) ≈ 7.348

So the missing side length is approximately 7.348. To find the other missing angles, we can use the fact that the angles in a triangle add up to 180°. Angle C is also 60°, so we can find angle A:

A + 60° + 60° = 180°
A = 60°

Therefore, all angles in the triangle are 60° and all sides are approximately 7.348.

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Answer:

Let's represent the number of stones each bag holds by the variable x.

Then the total number of stones in the 6 large bags is 6x, and the total number of stones in the 4 small bags is 4x.

Therefore, the total number of stones can be represented by the polynomial:

6x + 4x = 10x

So the polynomial is 10x, which represents the total number of stones.

Answer:

Step-by-step explanation:

large bags = x

small bags = y

6x + 4y

Answer:3.99

Step-by-step explanation:

i just kniow

From the inequality 3 < u < 2, we know that u is negative. Dividing both sides by 2, we get: 3/2 < u/2 < 1. So u/2 is also negative.  Negative values lie in Quadrants II and III. Since u/2 is between 3/2 and 1, it is closer to 1, which is the x-axis. Therefore, u/2 is in Quadrant III.

Given the inequality 3/2 < u < 2, we need to determine the quadrant in which u/2 lies.

First, let's find the range of u/2 by dividing the inequality by 2:
(3/2) / 2 < u/2 < 2 / 2
3/4 < u/2 < 1

Now, we can see that u/2 lies between 3/4 and 1. In terms of radians, this range corresponds to approximately 0.589 and 1.571 radians. This range falls within Quadrant I (0 to π/2 or 0 to 1.571 radians). Therefore, u/2 lies in Quadrant I.

Learn more about Quadrants here: brainly.com/question/7196312

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