Given a map with a scale of 1 inch = 10 feet, if a room is 15 feet wide, how wide should it appear on the map?​

The number of gigabytes that Mayerlin can use while staying within her budget is 5.

The first step is to determine the appropriate inequality sign to use. Here are the inequality signs:

> means greater than

< means less than

≥ means greater than or equal to

≤ less than or equal to

The sign that would be used is <

The form of the inequality is:

flat cost + (cost per gigabyte x number of gigabytes) < total budget

$55 + $5x < $70

5x < $70 - $55

5x < 25

x < 25 / 5

x = 5

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The number of black beads is 42, the number of white beads is 12, and there are 30 more black beads than white beads.

Mabel has a total of 54 decorative beads, some are black, and some are white. The ratio of the number of black beads to the number of white beads is 7:2. We have to find how many more black beads than white beads.

Let the total number of black beads be 7x, and that of white beads be 2x, respectively.

So, according to the problem, 7x + 2x = 54⇒ 9x = 54⇒ x = 6

Hence, the total number of black beads = 7x = 7 × 6 = 42

And the total number of white beads = 2x = 2 × 6 = 12

Therefore, the required difference between the number of black beads and white beads is 42 - 12 = 30.

Therefore, there are 30 more black beads than white beads.

: The number of black beads is 42, the number of white beads is 12, and there are 30 more black beads than white beads.

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Answer:

8.  x=19

9. 42

10. 23

Step-by-step explanation:

8.

x+x+4+3x-9 =90

5x-5=90

x= 19

9.

x=19

x+4= 23

19+23=42

10.

x+4=23

The voltage equation, we get:

v(t) = 140/29 + (√29/58)cos(√29t) + (√29/58)sin(√29t) + (3 - y)/29

Given that the differential equation is

d²v/dt² + 29v + y = 3,

and the initial conditions are

v(0) = 5 and dv/dt(0) = 1.

The characteristic equation is

m² + 29 = 0.

So, m₁ = i√29 and m₂ = -i√29.

Thus, the complementary function is vc

f(t) = c₁ cos (√29t) + c₂ sin (√29t)

where c₁ and c₂ are constants.

To determine the particular integral, we first determine the particular integral of y, which is a constant.

Since the right side of the equation is 3, we guess that the particular integral will be of the form y

p(t) = At² + Bt + C.

Substituting this into the differential equation, we get:

d²(At² + Bt + C)/dt² + 29(At² + Bt + C) + y

= 3 2Ad²t/dt² + 29At² + 58Bt + 29 C + y

= 3

Equating coefficients of t², t, and constants gives us:

2A + 29A = 0

⇒ A = 0, and

29C + y = 3

⇒ C = (3 - y)/29

The coefficient of t is 58B, which must equal 0 since there is no t term on the right side of the equation.

Thus, B = 0.

So, yp(t) = (3 - y)/29 is the particular integral of y.

Substituting this into the voltage equation, we get:

v(t) = D + c₁ cos (√29t) + c₂ sin (√29t) + (3 - y)/29

To determine the constants, we use the initial conditions:

v(0) = 5

⇒ D + (3 - y)/29 = 5

⇒ D = 140/29 dv/dt(0) = 1

⇒ -c₁√29 + c₂√29 = 1

From this, we get c₁ = c₂ = √29/58.

Finally, substituting all the values in the voltage equation,

v(t) = 140/29 + (√29/58)cos(√29t) + (√29/58)sin(√29t) + (3 - y)/29

Putting A = 0, B = 0, s3 = √29, and D = 140/29 in the voltage equation, we get:

v(t) = 140/29 + (√29/58)cos(√29t) + (√29/58)sin(√29t) + (3 - y)/29

where A = 0, B = 0, s3 = √29, and D = 140/29.

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The equation that represent the relationship of the coffees is 3x + 5y = 68.

At a coffee shop, two sizes of coffee are sold. A medium coffee costs $3, and a large coffee costs $5.

The equation that represent the relationship between the number of medium coffees sold, x, and the number of large coffees sold, y, in an hour where sales totalled $68 is as follows:

where

Therefore, the equation is as follows:

3x + 5y = 68

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Answer:

C. It is not a good fit because there are no points on the line.

Step-by-step explanation:

In order for a line to be a good fit for a data set represented as a scatterplot, the line must follow the general trend of the data in the scatterplot. This line does not follow the general trend of the data on the scatterplot, thus option (C) is the best statement to describe the situation.

C. It is not a good fit because there are no points on the line.

Answer:

solution,

\(f(x) = - {x}^{2} - 3x \\ f( - 10) = {-(10)}^{2} - 3 \times 10 \\ \: \: \: \: \: \: \: \: \: = -100 - 30 \\ \: \: \: \: \: = -130\)

hope this helps...

Good luck on your assignment..

Answer:

-130

Step-by-step explanation:

f(x) = -x^2 -3x

Let x= 10

f(10) = -(10)^2 -3*10

    = -100 -30

    = -130

When comparing variables x and y that are in the list '(AB), the comparison function we use is EQ.

Here's an explanation:

In Lisp, the EQ operator compares whether the two arguments are the same memory location.

EQV, on the other hand, checks whether the two values being compared are equivalent to each other.

For example, EQV will return true for two identical strings, while EQ will not.

The comparison function that should be used to compare variables x and y, which are both in the list '(AB), is EQ.

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Step-by-step explanation:

In triangle ABC. In triangle DBC

AB. = DB.

[Given]

<ABC. = <DBC

[Given]

BC. = BC

[Common side]

Hence,

As two sides and one angle of triangle ABC and triangle DBC are equal.

Triangle ABC is congruent to Triangle DBC

The optimal feasible solution is x1 = 1, x2 = 0.4, and the problem is a regular linear programming problem.

To solve the given linear programming problem, let's define the decision variables and formulate the objective function and constraints:

Decision Variables:

x1, x2

Objective Function:

Maximize: 2x1 + 6x2

Constraints:

2x1 + 5x2 ≤ 4

4x1 + x2 ≤ 6

x1, x2 > 0

To find the optimal feasible solution, we can use a linear programming solver. Here is the optimal solution for the given problem:

Optimal Solution:

x1 = 1

x2 = 0.4

The maximum value of the objective function is obtained when x1 = 1 and x2 = 0.4. The maximum value is 2(1) + 6(0.4) = 4.8.

Now let's analyze if the solution is a special case of linear programming.

This problem falls under the category of Linear Programming (LP) problems. However, it does not represent any specific special case of LP such as degeneracy, unboundedness, or infeasibility. The given problem has a feasible solution, and the objective function is maximized within the given constraints. Hence, it is a regular LP problem without any special case conditions.

Note: Since the solution is text-based, there is no need to scan or provide a separate file.

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(a) y = cx⁵ and 5x² + y² = k are do not satisfy the conditions for orthogonality.

(b) x² + y² = 2cy and x² + y² = 2kx potentially represent families of orthogonal functions .

(c) x² = y² + c and y = kx potentially represent families of orthogonal functions .

a. y = cx⁵ and 5x² + y² = k: Definitely incorrect. The equation 5x² + y² = k represents an ellipse, whereas the equation y = cx⁵ represents a polynomial function. These two functions do not satisfy the conditions for orthogonality.

b. x² + y² = 2cy and x² + y² = 2kx: Potentially correct. Both equations represent circles in the xy-plane. However, for them to be orthogonal functions, the centers of the circles need to coincide at the origin (0, 0). If the centers are at the origin, then these equations can be potentially correct as orthogonal functions.

c. x² = y² + c and y = kx: Potentially correct. The equation x² = y² + c represents a hyperbola, and the equation y = kx represents a straight line passing through the origin. If the hyperbola is centered at the origin and the line passes through the origin, then these equations can be potentially correct as orthogonal functions.

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In order to design a Class C amplifier with a 3-watt output and 99% efficiency, several considerations need to be taken into account, such as the choice of active device, biasing, impedance matching, and tuning.  

Designing a Class C amplifier with a specific output power and efficiency requires careful consideration of various parameters. Firstly, select a suitable active device, such as a transistor or a MOSFET, capable of handling the desired power level. The active device should have high gain and efficiency characteristics. Next, proper biasing of the active device is essential to ensure it operates in the Class C region. Biasing circuits, such as an LC or RC network, can be used to provide the necessary bias voltage and current. Impedance matching between the input and output circuits is crucial to maximize power transfer efficiency. Matching networks, consisting of inductors, capacitors, and transmission lines, can be used to match the amplifier's input and output impedances to the source and load impedances, respectively.

Tuning the amplifier involves adjusting the resonant frequency of the input and output circuits to optimize performance. This can be done using variable capacitors or inductors. Lastly, thorough testing and characterization of the amplifier should be performed to ensure it meets the desired specifications. This includes measuring power output, efficiency, distortion levels, and frequency response. It is important to note that designing a Class C amplifier with high efficiency and specific output power requires expertise in amplifier design and RF engineering. Working with experienced professionals or consulting relevant literature and resources can greatly assist in achieving the desired results.

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Question:

The graph shows the amount of money Miguel earns after working x hours. What is the rate of change of the amount earned with respect to hours worked for this function? hours per dollar hours per dollar 3 dollars per hour 13 dollars per hour

Answer: 13 dollars per hour

Step-by-step explanation:

The rate of change (ROC) for the point defined on the graph can be deduced by calculating the ratio of the difference between the two coordinate points given, similar to calculating the gradient or slope :

Point a = (5, 65) = (x2, y2)

Point b = (2, 26) = (x1, y1)

Calculating the gradient or slope :

Rate of change = (y2 - y1) / (x2 - x1)

Rate of change = (65 - 26) / (5 - 2)

Rate of change = 39 / 3

= 13 dollars per hour

Answer:

I think that the answer is D

Answer:

Step-by-step explanation:

19-12=7x3=21

The standard deviation of the sampling distribution of the sample mean would be approximately 2.8284

To determine the standard deviation of the sampling distribution of the sample mean, we will use the formula;

σ_mean = σ / √n

where σ is the standard deviation of the population that is 20 and n is the sample size (n = 50).

So,

σ_mean = 20 / √50 = 20 / 7.07

σ_mean  = 2.8284

The standard deviation of the sampling distribution of the sample mean is approximately 2.8284 it refers to that the sample mean would typically deviate from the population mean by about 2.8284, assuming that the sample is selected randomly from the population.

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The complete question is;

Another application of the sampling distribution of the sample mean Suppose that, out of a total of 559 full-service restaurants in Delaware, the number of seats per restaurant is normally distributed with mean mu = 99.2 and standard deviation sigma = 20. The Delaware tourism board selects a simple random sample of 50 full-service restaurants located within the state and determines the mean number of seats per restaurant for the sample. The standard deviation of the sampling distribution of the sample mean is Use the tool below to answer the question that follows. There is a.25 probability that the sample mean is less than

Answer:

4800

Step-by-step explanation:

→ Find how 15yr periods there are in 45 years

45 ÷ 15 = 3

→ Substitute this value as x into the equation

75 × 4³ = 4800

Answer:

Connor will earn more interest because he is getting compound interest.

Step-by-step explanation:

I have to assume in this case that the interest is 3.5% Per year:

Simple Interest = P(r + t)

{P = Principle, r = interest rate, t = time)

Wendy = $8,300 x (3.5% x 25)

Wendy = $7,262.50 (2 d.p)

Wendy = $8,300 + $7,262.50

Wendy = $15,562.50

Compound Interest = \(P (1 +r )^{t}\)

{P = Principle, r = interest rate, t = time)

Connor = $8,300 x \((1 + 0.035)^{25}\)

Connor = $19,614.93

Wendy's Interest = $15,562.50 - $8,300 = $7,262.50

Connor's interest = $ 19,614.93 - $8,300 = $11,314.93

Connor earned more interest by $4,052.43

___________________________________________________

Hope this helps!

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There would be about 12 pieces of fruit in the bag.

Step-by-step explanation:

since all sides of a cube have the same length (same principle as e.g. for a square), this means also the length and width are 10 in long.

the volume is therefore

10 × 10 × 10 = 1000 in³

Step-by-step explanation:

Aa = Ab × 10

that means that for the side lenghts

Sa² = Sb² × 10

Sa = Sb × sqrt(10)

Sa/Sb = sqrt(10)

Pa = 4×Sa

Pb = 4×Sb

Pa/Pb = 4×Sa / 4×Sb = Sa/Sb = sqrt(10) (= sqrt(10)/1)

the ratio of the perimeters is therefore

sqrt(10) : 1

Therefore, the rank of matrix 'a' in this case would be 5.

To find the rank of a matrix, we need to perform row reduction to obtain its row echelon form (REF) or reduced row echelon form (RREF). However, since the matrix 'a' is not provided, I cannot perform the calculations or determine its rank.

The rank of a matrix is equal to the number of non-zero rows in its row echelon form or reduced row echelon form. If the system of equations 'ax = 0' has a two-dimensional solution space, it means that the rank of matrix 'a' is less than the number of columns (6) but greater than 4 (since the solution space is two-dimensional).

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H(t) = -16t^2 +704t = -16(t^2 -44t)

Putting in vertex form: -16(t^2 -44t +484) + 704

h(t) = -16(t-22)^2 + 704

It takes 22 seconds to reach the maximum height, which is the vertex of the parabola at (22,704)

please mark me as the BRAINLIEST.....

Answer:

Step-by-step explanation:

Given function:

Maximum value is obtained at vertex. Using vertex formula to find the value of t at maximum height: x = -b/2a of a  quadratic formula ax² + bx + c

To solve this problem, we can use a system of equations. Let's denote the amount of 0.72 beans as x and the amount of 0.52 beans as y.

From the given information, we have two equations:
Equation 1: x + y = 110  (since 110 pounds of beans are used in total)
Equation 2: (0.72x + 0.52y) / 110 = 0.74  (since the mixture is worth 0.74 a pound)
To solve this system of equations, we can use the method of substitution. Let's solve Equation 1 for x:

x = 110 - y.
Now we substitute x in Equation 2 with 110 - y:
(0.72(110 - y) + 0.52y) / 110 = 0.74
Simplifying this equation, we get:
79.2 - 0.72y + 0.52y = 81.4
Combine like terms:
0.52y - 0.72y = 81.4 - 79.2
-0.20y = 2.2
Divide both sides by -0.20:
y = -2.2 / -0.20
y = 11
Substituting the value of y back into Equation 1:
x + 11 = 110
x = 110 - 11
x = 99
Therefore, 99 pounds of 0.72 beans and 11 pounds of 0.52 beans are used in the mixture.
99 pounds of 0.72 beans and 11 pounds of 0.52 beans are used in the mixture.
To solve this problem, we used a system of equations to represent the given information. We assigned variables to the amounts of 0.72 beans and 0.52 beans used in the mixture.

By setting up two equations based on the total weight of the beans and the average price per pound, we were able to solve for the values of x and y.

Using the method of substitution, we substituted the value of x in Equation 2, simplified the equation, and solved for y. Then, we substituted the value of y back into Equation 1 to solve for x.

The final solution showed that 99 pounds of 0.72 beans and 11 pounds of 0.52 beans were used in the mixture.
The solution we obtained is reasonable because it satisfies both equations and meets the given conditions.

The total weight of the beans (99 + 11 = 110) matches the given total weight, and the average price per pound

(0.72 × 99 + 0.52 × 11) / 110 = 0.74 matches the given average price.

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Answer:

x/20=15/12,so x =25

Step-by-step explanation:

the awnser is d (6.-3) and -1,4