hi, pls help.
Why do Jews consider the Western Wall in Jerusalem a sacred place? *
1. It was built by King Solomon.
2. It was all that was left of the Second Temple after the Romans destroyed it.
3. It was the place where Abraham settled.
4. It was constructed with the stones that held the Ten Commandments.

Answer:

8.9

Step-by-step explanation:

Answer:

n <−4

42 - 28n + 4n >138

42 - 24n >138

-24n > 138 - 42

-24n >96

n <−96/24

N <−4

Answer:  

c

Step-by-step explanation:  

Answer:

1/4

Step-by-step explanation:

Answer:

$6

Step-by-step explanation:

Answer:

Step-by-step explanation:

Just plug each value of x into the equation to get y:

f(-2) = (√-(-2) - 2)  + 2 = √0 + 2 = 2

f(-3) = (√-(-3) - 2) + 2 = √1 + 2 = 3

f(-6) = (√-(-6) - 2) + 2 = √4 + 2 = 4

f(-11) = (-(-11) - 2) + 2 = √9 + 2 = 5

The slope of the simple linear regression equation represents the average change in the value of the dependent variable per unit change in the independent variable (x).

A linear regression equation is the formula for the straight line that best represents a given dataset in statistics. The equation represents the relationship between the dependent and independent variables with the help of a straight line.

It is often used to predict or forecast the dependent variable values based on the independent variable values.A slope is a measure of the steepness of the line in the linear regression equation.

It refers to the rate of change of the dependent variable concerning the independent variable.

The slope of the equation is denoted by the symbol “m”.In conclusion, the slope of the simple linear regression equation represents the average change in the value of the dependent variable per unit change in the independent variable (x).

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Answer:

(a). and (b).

Step-by-step explanation:

Simplify Square Roots -

Find one perfect square and one imperfect square to multiply to the square root.

\( \sqrt{4 \times 10} = 2 \sqrt{10} \)

\(40^{ \frac{1}{2} } = \sqrt[2]{(40) ^{1} } = \sqrt{40} \)

\( \sqrt{40} \neq 4 \sqrt{10} \)

\(160 ^{ \frac{1}{2} } = \sqrt[2]{160 ^{1} } = \sqrt{160} \neq \: \sqrt{40} \)

\(5 \sqrt{8} \neq \: \sqrt{40} \)

To show that a tree has at most   \(n/2\)many vertices that have degree 3 or higher, we will use proof by contradiction.

Assume that there exists a tree with more than \(n/2\) vertices that have a degree of 3 or higher. Let V be the set of vertices in the tree, and let \(V_3\) be the set of vertices in V that have a degree 3 or higher. Let k be the number of vertices in\(V_3.\)

Since the tree has n vertices, there are n-k vertices in V that have degree 1 or 2. Since each vertex in the tree has degree at least 1, we have \(n-k ≤ n\), which implies that k ≥ 0.

Now, consider the sum of degrees of all vertices in the tree. By definition of a tree, this sum is twice the number of edges in the tree, which is n-1. Therefore, we have:

\(2(n-1) = Σ_degrees\)

where \(Σ_degrees\) is the sum of degrees of all vertices in the tree.

Let d_i be the degree of the i-th vertex in V_3. Since each vertex in V_3 has degree 3 or higher, we have d_i ≥ 3 for all i. Therefore, the sum of degrees of vertices in V_3 is at least 3k.

Let m be the number of vertices in V that have degree 1 or 2. Let d_j be the degree of the j-th vertex in V that has degree 1 or 2. Since each vertex in V that has degree 1 or 2 has degree at most 2, we have d_j ≤ 2 for all j. Therefore, the sum of degrees of vertices in V that have degree 1 or 2 is at most 2m.

Since V is the disjoint union of \(V_3\) and the set of vertices in V that have degree 1 or 2, we have:

\(Σ_degrees = Σ_{i=1}^k d_i + Σ_{j=1}^m d_j\)

Combining the inequalities \(3k ≤ Σ_{i=1}^k d_i and Σ_{j=1}^m d_j ≤ 2m, we get:\\Σ_degrees ≥ 3k + Σ_{j=1}^m d_j ≥ 3k\)

where the last inequality follows from for all j.

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A direct relationship between two variables is reflected in a "POSITIVE" correlation coefficient.

Correlation is a statistical technique for measuring and describing the relationship between two variables.

The variables move in the same direction when they have a positive correlation. In other words, as one variable increases, so does the other, and conversely, as one variable decreases, so does the other.

Typically, the two variables are simply observed rather than manipulated. Two scores from the same individuals are required for the correlation.

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The rate at which the water level in reservoir 1 drops can be calculated as follows:

Let the volume of reservoir 1 be V1, and the surface area be A1. The volume of water that is flowing out of reservoir 1 per minute is given by;

V = A1 x h

where h is the height of water that flows out of reservoir 1 per minute. Since the water level in reservoir 1 drops at a rate of 0.01 m/min, then the height of water that flows out of reservoir 1 per minute is 0.01 m/min. Therefore, the volume of water that flows out of reservoir 1 per minute is given by;

V = A1 x hV = A1 x 0.01 .

Assuming that the cross-sectional area of the pipe that connects the two reservoirs is the same as that of reservoir 1 (A1), then the volume of water that flows into reservoir 2 per minute is also given by;

V = A2 x h

where A2 is the surface area of reservoir 2.

Therefore, the rate at which the water level in reservoir 2 rises can be calculated as follows:

Let the volume of reservoir 2 be V2. The volume of water that flows into reservoir 2 per minute is given by;

V = A2 x hA1 x 0.01 = A2 x h = (A1 / A2) x 0.01

Since the height of water that flows out of reservoir 1 per minute is 0.01 m/min, then the height of water that flows into reservoir 2 per minute is (A1 / A2) x 0.01. Therefore, the rate at which the water level in reservoir 2 rises is given by;

(A1 / A2) x 0.01 m/min.

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The median and quartiles fall in the middle age classes.

The median is the middle value in a set of data, meaning that half of the data falls below the median and half falls above it. The quartiles divide the data into four equal parts, with the first quartile (Q1) being the 25th percentile, the second quartile (Q2) being the median or 50th percentile, and the third quartile (Q3) being the 75th percentile.

In terms of age classes, the median and quartiles would fall in the middle age classes. For example, if the age classes were 0-10, 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, and 81-90, the median and quartiles would fall in the 21-30, 31-40, and 41-50 age classes.

It is important to note that the specific age classes that the median and quartiles fall in will depend on the distribution of the data. However, they will always fall in the middle age classes, as they represent the middle values of the data set.

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a) The amount of water in the pool after 2 hours can be calculated using the equation.

Water in pool = 10L/hour × 2 hours = 20L.

b) The pool will be 80L when the equation is satisfied: 80L = 10L/hour × Time.

Solving for Time, we find Time = 8 hours.

c) Part b) is linear.

a) To calculate the amount of water in the pool after 2 hours, we can use the equation:

Water in pool = Water filling rate × Time

Since the pool gets filled up with 10L of water per hour, we can substitute the values:

Water in pool = 10 L/hour × 2 hours = 20L

Therefore, after 2 hours, there will be 20 liters of water in the pool.

b) To determine the number of hours it takes for the pool to reach 80 liters, we can set up the equation:

Water in pool = Water filling rate × Time

We want the water in the pool to be 80 liters, so the equation becomes:

80L = 10 L/hour × Time

Dividing both sides by 10 L/hour, we get:

Time = 80L / 10 L/hour = 8 hours

Therefore, it will take 8 hours for the pool to contain 80 liters of water.

c) Part b) is linear.

The equation Water in pool = Water filling rate × Time represents a linear relationship because the amount of water in the pool increases linearly with respect to time.

Each hour, the pool fills up with a constant rate of 10 liters, leading to a proportional increase in the total volume of water in the pool.

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Answer:

6 weeks

Step-by-step explanation:

Let x represent the weeks

\(25=10+ (10x-7.5x = 2.5)\)

\(25-10=2.5x = 15\)

\(\frac{15}{2.5}=x \\(150/25)\\x=6\)

Carol and Cathy will have same saved amount after 6 weeks.

What do you mean by algebraic expressions ?

Algebraic expression is an equation which consists of variables and the arithmetic operations such as division , multiplication , etc.

It is given that Carol started with $25 and saved per week an amount of $7.50 whereas Cathy started with $10 and saved per week amount of $10.

Let's assume that after x no. of weeks they will have same saved amount.

So , we can write an algebraic expression which is :

25 + (7.5 × x) = 10 + (10 × x)

Now , reordering the like terms we get :

10x - 7.5 x = 25 - 10

Simplifying this we get :

2.5 x = 15

or

x = 15 ÷2.5

or

x = 6 weeks

Therefore , Carol and Cathy will have same saved amount after 6 weeks.

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Answer:

y= 2x - 1

Step-by-step explanation:

i hope that helps

Answer:

n = 2m + 1

Step-by-step explanation:

\(-1=2m-n\\\\-1+n=2m-n+n\\\\-1+n=2m\\\\-1+1=n=2m+1\\\\\boxed{n=2m+1}\)

Hope this helps.

Jenna and Mia arrived at the same answer because both the methods they have adopted are correct.

Given solution of a question done by two people.

Jenna's method

5(30+4)

(5)(30)+(5)(4)

150+20=170

Mia's method

5(30+4)

5(34)=170

We are required to find why they have achieved at the same answer.

Jenna and Mia have achieved at the same answer because they both have adopted the right method.Jenna's method is the equation in its simplest form. Addition of that is easily understood. While Mia's method is just valid as Jenna's some people may have trouble remembering the steps of multiplication within parantheses as well as being able to look at large multiplication problem and automatically knowing the answer or being able to get the answer as fast as simple addition.

Hence Jenna and Mia arrived at the same answer because both the methods they have adopted are correct.

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Answer:

\(\frac{3}{4}\) or 0.75

Step-by-step explanation:

I apologize if I'm incorrect

TRUE: Basketball players' height is seen as a continuous variable.

Yes, as lengths are continuous variables, the random variable does indeed refer to length.

Thus, Basketball players' height is seen as a continuous variable

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A measurement that is closest to the area of the tabletop in square feet is 50 ft².

Mathematically, the circumference of a circle can be calculated by using this mathematical expression:

C = 2πr or C = πD

Where:

Substituting the given parameters into the circumference of a circle formula, we have the following radius;

8π = 2πr

r = 8π/2π

Radius, r = 4 feet.

Mathematically, the area of a circle can be calculated by using this formula:

Area of a circle = πr²

Area of a circle = 3.14(4)²

Area of a circle = 50.24 ≈ 50 ft².

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Answer:

James is correct.

Step-by-step explanation:

We want to see if -1*n is equivalent to n/(-1).

We will see that these operations are equivalent, so Jaime is correct.

So, let's start with the original equation:

-1*n

Notice that we can rewrite:

-1 = -1/1 = -(1/1) = 1/(-1)

Then if we write:

-1*n = (1/(-1))*n = (1*n)/(-1) = n/(-1)

So Jaime is actually correct, this happens because when we multiply or divide by -1 we do not change the value of n, we only change the sign, so these two operations are equivalent.

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Answer:

a+b+c=59

Step-by-step explanation:

a+b=29

b+c=45

a+c=44

Therefore we know that a and b are each equal to less than 29. And c must be equal to less than 44. By messing around with those numbers I got that if a is equal to 14 and b is equal to 15, that gives us 29. And if b is equal to 15, then 15 minus 45 is equal to 30. Meaning that c would equal 30. And if we add c + a, or 30+14, that is equal to 44. So then a+b+c=? can be written as 14+15+30=59

Answer:

The jelly beans cost $2.25 / pound

The trail mix costs $1.25 / pound

Step-by-step explanation:

Let +a+ = cost/pound of jelly beans

Let +b+ = cost/pound of trail mix

(1) 6a + 2b =16

(2) 3a + 5b =13

Multiply both sides by 2  

and subtract (1) from (2)

(2) 6a -2b = -16

(1) -6a -2b = -16

8b=10

b=1.25

and

(1) 6a+2b=16

(1) +3a+b=8

(1) 3a+1.25=8

(1) +3a=6.75

(1) a=2.25

Answer:

b) f(x)=(x+4)(x-2) 2.....

The zeros on the graph satisfy the function f(x)=(x+4)(x-2)². Therefore, option B is the correct answer.

In the given graph, the function is plotted.

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

The zero of a polynomial is (2, 0) and (-4, 0)

A. f(x)=(x-4)(x+2)²

Put (2, 0) in y=(x-4)(x+2)², we get 0=(2-4)(2+2)²

LHS≠RHS

B) f(x)=(x+4)(x-2)²

Put (2, 0) in y=(x+4)(x-2)², we get 0=(2+4)(2-2)²

LHS=RHS

Put (-4, 0) in y=(x+4)(x-2)², we get 0=(-4+4)(-4-2)²

LHS=RHS

C) f(x)=(x+4)²(x-2)

Put (2, 0) in y=(x+4)²(x+2), we get 0=(2+4)²(2+2)

LHS≠RHS

D) f(x)=(x-4)²(x+2)

Put (2, 0) in y=(x-4)²(x+2), we get 0=(2-4)²(2+2)

LHS≠RHS

The zeros on the graph satisfy the function f(x)=(x+4)(x-2)². Therefore, option B is the correct answer.

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The correct answer is A.\(\((5 \frac{1}{\overline{9}}) \times (-0.\overline{3})\)\), as it involves the multiplication of two rational numbers, resulting in a rational number.

Therefore, the product of these two rational numbers in option A will yield a rational number.

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The derivative of a function represents the rate of change of the function with respect to its variable. This rate of change is described as the slope of the tangent line to the curve of the function at a specific point. The derivative of the cosine function can be found by applying the limit definition of the derivative to the cosine function.

\(f(x) = cos(x) then f'(x) = -sin(x)\).

Let's proceed with the proof.  Definition of the Derivative: The derivative of a function f(x) at x is defined as the limit as h approaches zero of the difference quotient \(f(x + h) - f(x) / h\) if this limit exists. Using this definition, we can find the derivative of the cosine function as follows:

\(f(x) = cos(x) f(x + h) = cos(x + h)\)

Now, we can substitute these expressions into the difference quotient: \(f'(x) = lim h→0 [cos(x + h) - cos(x)] / h\)

We can then simplify the expression by using the trigonometric identity for the difference of two angles:

\(cos(a - b) = cos(a)cos(b) + sin(a)sin(b)\)

Applying this identity to the numerator of the difference quotient, we obtain:

\(f'(x) = lim h→0 [cos(x)cos(h) - sin(x)sin(h) - cos(x)] / h\)

We can then factor out a cos(x) term from the numerator:

\(f'(x) = lim h→0 [cos(x)(cos(h) - 1) - sin(x)sin(h)] / h\)

We can then apply the limit laws to separate the limit into two limits:

\(f'(x) = lim h→0 cos(x) [lim h→0 (cos(h) - 1) / h] - lim h→0 sin(x) [lim h→0 sin(h) / h]\)

The first limit can be evaluated using L'Hopital's rule:

\(lim h→0 (cos(h) - 1) / h = lim h→0 -sin(h) / 1 = 0\)

Therefore, the first limit becomes zero:

\(f'(x) = lim h→0 - sin(x) [lim h→0 sin(h) / h]\)

Applying L'Hopital's rule to the second limit, we obtain:

\(lim h→0 sin(h) / h = lim h→0 cos(h) / 1 = 1\)

Therefore, the second limit becomes 1:

\(f'(x) = -sin(x)\)

Thus, we have proved that if \(f(x) = cos(x), then f'(x) = -sin(x)\).

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