-4-2/3x= -6 by adding 4 to both sides

Answer:

Option 3

Step-by-step explanation:

In the 3rd graph, as it passes through the median of the trend of the data, we can consider to be the line of best fit.

Answer:

6(x^2-xy) - 3x(x-2) = 6x^2 - 6xy - 3x^2 + 6x

Combining like terms: 3x^2 - 6xy + 6x

To express the result as a monomial it would be 3x^2

So, the final answer is 3x^2

The x component is -4 and the y component is \((3√2)/2 + (3√2)/2 = 3√2.\)
Adding -4 and 3√2, we get \(-4 + 3√2.\) Comparing the total displacements of Group A and Group B, we can determine which group is closer to the lodge.

In problem 2A, if Group A goes 4 miles west and then turns 45⁰ north of west and travels 3 miles, we can use vector addition to determine the displacement.
First, we need to break down the displacement into its x and y components. Going 4 miles west means moving -4 miles in the x-direction.

Turning 45⁰ north of west means moving in a diagonal direction, which we can split into its x and y components.

To find the x component, we can use cosine of 45⁰, which is \((√2)/2\).

So, the x component would be\((√2)/2 * 3 = (3√2)/2.\)

To find the y component, we can use sine of 45⁰, which is \((√2)/2\). So, the y component would be \((√2)/2 * 3 = (3√2)/2.\)

Now, we can add the x and y components to find the total displacement. The x component is -4 and the y component is \((3√2)/2 + (3√2)/2 = 3√2.\)
Adding -4 and 3√2, we get \(-4 + 3√2.\)

Comparing the total displacements of Group A and Group B, we can determine which group is closer to the lodge.

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Group B is closer to the lodge.

In problem 2A, Group A initially goes 4 miles west. Then, they turn 45 degrees north of west and travel 3 miles. To determine which group is closer to the lodge, we need to compare the final positions of the two groups.

Group B initially moves 5 miles west. Since Group A traveled 4 miles west, Group B is 1 mile farther from the lodge at this point.

Next, Group A turns 45 degrees north of west and travels 3 miles. We can break this motion into its north and west components. The north component is 3 * sin(45) = 2.12 miles, and the west component is 3 * cos(45) = 2.12 miles.

To find the final position of Group A, we add the north component to the initial north position (0 miles) and the west component to the initial west position (4 miles). Therefore, Group A's final position is at 2.12 miles north and 6.12 miles west.

Comparing the final positions, Group A is closer to the lodge. The distance from the lodge to Group A is sqrt((0-2.12)^2 + (5-6.12)^2) = 2.12 miles. The distance from the lodge to Group B is sqrt((0-0)^2 + (5-4)^2) = 1 mile. Therefore, Group B is closer to the lodge.

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Answer:

p= -16.1

Step-by-step explanation:

-4.8=0.5(p+6.5)

first you want to multiply 0.5 by p and 6.5.

-4.8=0.5(p)+0.5(6.5)

then solve it:

-4.8=0.5p+3.25

next you want to isolate p so you'd subtract 3.25 on both sides so:

(-4.8-3.25)=0.5p+(3.25-3.25)

solve:

-8.05=0.5p

lastly, divide 0.5 on both sides:

-16.1=p or p= -16.1

hope this helps :)

Answer:

I only know number 7 sorry

Step-by-step explanation:

Anyways.....

The solution she has in the bucket has 80% water (as you found out), including 14.4 fl oz. water in a total of 18 fl oz. solution.

Barbara will need to add ammonia, but only after removing some (x fl oz.) of the solution she has in the bucket.

If Barbara wants an 18 fl oz. bucket full of 25% ammonia, she will need to have in that bucket 18 × 0.25=4.5 fl oz. ammonia (25% ammonia) and

18 × 0.75=13.5 fl oz. water (75% water).

Barbara will have to start by removing some of that 80% water solution to end up with just 13.5 fl oz. water in the bucket. Removing x fl oz. of solution, gets rid of 0.8x fl oz. of water, leaving 14.4 - 0.8x fl oz.  of water. since the amount of water needed is 13.5 fl oz. Our equation is

13.5 = 14.4 - 0.8x

13.5= 14.4 - 0.8x →→ 0.8x = 14.4 - 13.5 →→ 0.8x = 0.9 →→ \(x= \frac{0.9}{0.8}\) →→ x=1.125 fl. oz

Therefore 1.125 (may need to round) fl. oz should be replaced with ammonia.

Hope this helps some of your stress.

Answer:

A, D and F

Step-by-step explanation:

cus they are an acute triangle because they are smaller than 90 degrees

Answer:

3/16

Step-by-step explanation:

you do length x width x height, and get 3/16

Mr. Steiner will need to pay $18774 in total and he paid $4774 more than the actual price of the car.

we know that the amount of money earned, in compound interest after t years is showed as, A(t) = P (1+\(\frac{r}{n}\))ⁿ\(^{t}\)

here, we know that A(t) is the amount of money after t years.

P is the initial sum of money, know as principal,

r is the interest rate as a decimal value,

n is the number of times that interest is compounded per year,

and

t is the time in years for which the money is invested or borrowed.

now here we know,

P = 14000, r = 0.049, n = 12, t = 6

when we have all values we can find,

A(t) = P(1+\(\frac{r}{n}\))ⁿ\(^{t}\)

A(6) = 14000 (1+\(\frac{0.049}{12}\))¹²ˣ⁶

A(6) = 18774

therefore we know that Mr. Steiner needs to pay $18774 in total.

now,

18774 - 14000 = 4774

therefore we get to know that Mr. Steiner paid $4774 more than the actual price of the car.

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a.  it implies that the matrix A - XI is not invertible. b)  the determinant of A - XI is zero. c) the equation: .2A - 1 = 0 d) the eigenvalue of matrix A is 5. e)  the eigenvalue of matrix A is 2. f) the eigenvalues of matrix A are X = 0 and X = 1.

a. If there is a nonzero solution to the homogeneous equation (A - XI)v = 0, where X is an eigenvalue of the matrix A, it implies that the matrix A - XI is not invertible. This is because for a matrix to be invertible, its determinant must be nonzero.

If there exists a nonzero solution to the homogeneous equation, it means that the determinant of A - XI is zero, indicating that A - XI is singular and not invertible.

b. If there is a nonzero solution to the homogeneous equation (A - XI)v = 0, it implies that the determinant of A - XI is zero. This is because the homogeneous equation represents a system of linear equations, and the determinant of the coefficient matrix (A - XI) being zero implies that the system has a nontrivial solution. Therefore, we can conclude that the determinant det(A - XI) must be zero.

c. Given the matrix A - AI = [1 2 2 1] - A [1 0 0 1] = ['- .2 1 -A'], to find the determinant det(A - 1), we substitute the value X = 1 into the matrix A - XI and compute its determinant. Evaluating the determinant, we have:

det(A - XI) = det(['- .2 1 -A']) = (-.2)(-A) - (1)(1) = .2A - 1

Setting this determinant equal to zero, we obtain the equation:

.2A - 1 = 0

d. Using the equation .2A - 1 = 0 obtained from the previous part, we solve it to find the eigenvalues by setting the determinant det(A - XI) = 0:

.2A - 1 = 0

.2A = 1

A = 1/.2

A = 5

Therefore, the eigenvalue of matrix A is 5.

e. For the matrix A = [2 0 1 2], we can find its eigenvalues by solving the equation det(A - XI) = 0:

det(A - XI) = det([2 0 1 2] - X [1 0 0 1]) = det([2-X 0 1 2-X])

Expanding the determinant, we have:

(2-X)(2-X) - (0)(1) = 0

(2-X)^2 - 0 = 0

(2-X)^2 = 0

Taking the square root, we get:

2-X = 0

X = 2

Therefore, the eigenvalue of matrix A is 2.

f. For the matrix A = [0 1 0 1], we can find its eigenvalues by solving the equation det(A - XI) = 0:

det(A - XI) = det([0 1 0 1] - X [1 0 0 1]) = det([-X 1 0 1-X])

Expanding the determinant, we have:

(-X)(1-X) - (1)(0) = 0

X^2 - X = 0

Factoring out X, we get:

X(X - 1) = 0

Therefore, the eigenvalues of matrix A are X = 0 and X = 1.

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Answer:

The balance of the savings account in 15 years will be $1,750.24.

Answer:

Step-by-step explanation:

y=10

Answer:

x = 3.2

Step-by-step explanation:

Since both polygons are similar, therefore their corresponding lengths would be proportional to each other.

Thus,

\( \frac{QP}{KJ} = \frac{TP}{NJ} \)

\( \frac{x}{4} = \frac{4}{5} \)

Multiply both sides by 4

\( \frac{x}{4}*4 = \frac{4}{5}*4 \)

\( x = \frac{4*4}{5} \)

x = 3.2

Answer:

2.7 trust me

Step-by-step explanation:

Answer:

14

Step-by-step explanation:

Answer:

x^3+8

Step-by-step explanation:

used an online calculator lol

Answer: BC/FG and AE/AC

Step-by-step explanation:

Answer:

50.24 in

Step-by-step explanation:

Answer:

(-3)² = 9

Step-by-step explanation:

\((-3)^{-2} = \frac{1}{9}\)

Answer:

1/9

Step-by-step explanation:

(-3) ^ -2

The negative exponent means  a^-b = 1/a^b

1/(-3)^2

1/(-3*-3)

1/9

Answer:

8b + 6.17 = 59.93

Step-by-step explanation:

if b= the cost of each bag of onions and your buying 8 then 8b represents the cost of 8 bags of onions and you bought one things of milk so you add 6.17 With the 8b, and your given the total, so set 8b+ 6.17 equal to the total

Answer:

I think its true although Im not sure

Answer:

y = -\(\frac{3}{4}\)x + 1

Step-by-step explanation:

y = mx + b

Find two point on the graph (0,1) (4,-2); 1 is the y intercept

y = mx + 1

\(\frac{1 + 2}{0 - 4}\) = \(\frac{3}{-4}\)

y = -\(\frac{3}{4}\)x + 1

Answer:

y= -3/4 + 1

Step-by-step explanation:

y= -3/4 + 1

Answer:

d = - \(\frac{1}{4}\)

Step-by-step explanation:

the common difference d is the difference between consecutive terms , so

d = a₂ - a₁ = \(\frac{1}{4}\) - \(\frac{1}{2}\) = - \(\frac{1}{4}\)

Component 3 must be started on day 197 to meet the delivery date of day 200.

To illustrate the backward schedule, we need to start from the delivery date (day 200) and work our way backward, taking into account the lead times and dependencies of each operation.

(a) Backward schedule:

Operation    |  Work Center  |  Time (hours)  |  Start Day

--------------------------------------------------------

Final Assembly |   Work Center 1  |        1       |     200

Sub-assembly 1 |   Work Center 2  |        2       |     199

Component 4     |   Work Center 3  |        3       |     197

Component 2     |   Work Center 4  |        4       |     196

Component 3     |   Work Center 5  |        3       |     ????

(b) To identify when component 3 must be started to meet the delivery date, we need to consider its dependencies and lead times.

From the backward schedule, we see that component 3 is required for sub-assembly 1, which is scheduled to start on day 199. The time required for sub-assembly 1 is 2 hours, which means it should be completed by the end of day 199.

Since component 3 is needed for sub-assembly 1, we can conclude that component 3 must be started at least 2 hours before the start of sub-assembly 1. Therefore, component 3 should be started on day 199 - 2 = 197 to ensure it is completed and ready for sub-assembly 1.

Hence, component 3 must be started on day 197 to meet the delivery date of day 200.

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2y = x  -8 is the equation of line w.

A linear equation is an algebraic equation of the form y=mx+b, where m is the slope and b is the y-intercept, and only a constant and a first-order (linear) term are included. Sometimes, the aforementioned is referred to as a "linear equation of two variables," with y and x serving as the variables.

Given, Line v has an equation of y= – 2x+5.

also given Perpendicular to line v is line w

Slope of the perpendicular line w = -1/(Slope of the line v)

From Slope intercept form

y = mx + c

where m is the slope

Slope of the perpendicular line w = -1/ (-2)

The slope of the perpendicular line w = 1/2

Thus, the equation of a line:

y = 1/2x + c

Since, w passes through the point (6, – 5).

Thus,

-5 = 1/2 * 6 + c

c = -8

So, the equation of line y = 1/2x -8

Therefore, the Equation of line w is 2y = x  -8.

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The value of the error for Z, where x = 5.9 +/- 0.5 and y = 2.1 +/- 0.2, with one decimal place is 4.

To calculate the error in Z, we need to consider the uncertainties in both x and y. The error in Z can be determined by propagating the uncertainties using the formula for error propagation.

In this case, Z is given by the equation Z = x/y. To propagate the uncertainties, we use the formula for relative error:

ΔZ/Z = sqrt((Δx/x)^2 + (Δy/y)^2)

Given the uncertainties Δx = 0.5 and Δy = 0.2, and the values x = 5.9 and y = 2.1, we substitute these values into the formula:

ΔZ/Z = sqrt((0.5/5.9)^2 + (0.2/2.1)^2) = sqrt(0.0089 + 0.0181) ≈ 0.134

Multiplying this value by 100 to convert it to a percentage, we get approximately 13.4%. Rounding to one decimal place, the value of the error is 4.

Therefore, the value of the error for Z, with one decimal place, is 4.

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The statement which represents the expression one over seven raised to the third power is Option (C) one over three hundred forty-three is the correct option.

We are given the expression:

one over seven raised to the third power

This can be written as:

(1 / 7) ³

Simplifying and solving the expression by finding the cube of 1 and 7.

1³ = 1

7³ = 343

So, the expression will now become:

(1 / 7) ³ = 1 / 343

which can also be written as:

one over three hundred forty - three

Therefore, option (C) one over three hundred forty-three is the correct option which represents the expression one over seven raised to the third power.

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"From a table of integrals, we know that for \(\(a \neq 0\)\) and \(\(b \neq 0\):\)

\(\[\int \cos(at) \, dt = \frac{1}{a} \cdot \cos(at) + \frac{1}{b} \cdot \sin(bt) + C\]\)

and

\(\[\int e^a t \cos(bt) \, dt = \frac{e^{at}}{a} \cdot \cos(bt) + \frac{b}{a^2 + b^2} \cdot \sin(bt) + C\]\)

Use this antiderivative to compute the following improper integral:

\(\[\int_{-\infty}^{0} \cos(3t) \, dt = \lim_{{T \to \infty}} \int_{0}^{T} e^t \cos(3t) \, e^{-st} \, dt = \lim_{{T \to \infty}} \text{ if } s \neq 1, \, \text{ or } \lim_{{T \to \infty}} \text{ if } s = 1.\]\)

For which values of \(\(s\)\) do the limits above exist? In other words, what is the domain of the Laplace transform of \(\(\frac{1}{\cos(3)} \cdot e^t \cos(3t)\)\)?

Evaluate the existing limit to compute the Laplace transform of  on the domain you determined in the previous part:

\(\[L\{e^t \cos(3t)\\).

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The intervals for which the function y = -7 - |x| is increasing and decreasing is given as follows:

The function is defined as follows:

y = -7 - |x|.

The parent absolute value function is increasing for x > 0 and decreasing for x < 0.

For this problem, the multiplication by -1 means that the function was reflected over the x-axis, meaning that the intervals of decrease and increase were exchanged, and thus they are given as follows:

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