1) (28 ÷ 4) + 3 + (10 - 8) × 5 2) 12 - 5 + 6 × 3 + 20 ÷ 4 3) 36 ÷ 9 + 48 - 10 ÷ 2 4) 10 + 8 × 90 ÷ 9 - 4 5) 8 × 3 + 70 ÷ 7 – 7

Answer:

i think you forgot to add how little he wants to spend

Step-by-step explanation:

Reuben wants the total number of dollars he spends on admission and rides to be fewer than ??? whats the number that he wants to spend

To solve the given differential equation:

y(ln(x) - ln(y)) dx = (x ln(x) - x ln(y) - y) dy

We can start by rearranging the terms:

y ln(x) dx - y ln(y) dx = x ln(x) dy - x ln(y) dy - y dy

Next, we can integrate both sides of the equation:

∫ y ln(x) dx - ∫ y ln(y) dx = ∫ x ln(x) dy - ∫ x ln(y) dy - ∫ y dy

To integrate the left-hand side, we can use integration by parts. Let's denote u = ln(x) and dv = y dx. Then, du = (1/x) dx and v = xy. Applying integration by parts, we have:

∫ y ln(x) dx = xy ln(x) - ∫ (1/x)(xy) dx

= xy ln(x) - ∫ y dx

= xy ln(x) - yx + C1

where C1 is the constant of integration.

Similarly, integrating the other terms:

∫ y ln(y) dx = xy ln(y) - yx + C2

∫ x ln(x) dy = (x^2 ln(x))/2 - ∫ (x^2)(1/x) dy

= (x^2 ln(x))/2 - ∫ x dy

= (x^2 ln(x))/2 - (x^2)/2 + C3

∫ x ln(y) dy = (x^2 ln(y))/2 - ∫ (x^2)(1/y) dy

= (x^2 ln(y))/2 - ∫ (x^2/y) dy

= (x^2 ln(y))/2 - x^2 ln(y) + ∫ x dy

= (x^2 ln(y))/2 - x^2 ln(y) + (x^2)/2 + C4

∫ y dy = (y^2)/2 + C5

Substituting these results back into the original equation:

xy ln(x) - yx + C1 - xy ln(y) + yx - C2 = (x^2 ln(x))/2 - (x^2)/2 + C3 - (x^2 ln(y))/2 + x^2 ln(y) - (x^2)/2 + C4 - (y^2)/2 - C5

Simplifying:

xy ln(x) - xy ln(y) = (x^2 ln(x))/2 - (x^2 ln(y))/2 - (y^2)/2 + C

where C = C1 - C2 + C3 + C4 - C5.

We can further simplify this equation:

xy (ln(x) - ln(y)) = (x^2 ln(x) - x^2 ln(y) - y^2)/2 + C

Finally, dividing both sides by (ln(x) - ln(y)), we get:

xy = (x^2 ln(x) - x^2 ln(y) - y^2)/(2(ln(x) - ln(y))) + C

This is the general solution to the given differential equation.

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The number that has no composite factors is (c) 77

From the question, we have the following parameters that can be used in our computation:

Number = 21

The factors of 21 are 3 and 7

These factors are composite numbers because they are prime numbers

using the above as a guide, we have the following:

The number 77 has no composite factors

This is so because

77 = 7 * 11

These factors are composite numbers because they are prime numbers

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there will be 27 zeros at the end of 115!.

To determine the number of zeros at the end of 115!, we need to consider the prime factorization of the number and examine how many factors of 5 are present.

A zero at the end of a factorial occurs when there is a factor of 10 present, which is equivalent to having both factors of 2 and 5. Since the number of factors of 2 is usually abundant, the crucial factor is the number of factors of 5.

In the prime factorization of 115!, the factors of 5 arise from the multiples of 5 (5, 10, 15, 20, ...) as well as higher powers of 5 (25, 50, 75, ...). We need to determine how many multiples of 5, multiples of 25, multiples of 125, and so on are present.

1. Multiples of 5: The number of multiples of 5 in 115! is given by ⌊115/5⌋ = 23.

2. Multiples of 25: The number of multiples of 25 in 115! is given by ⌊115/25⌋ = 4.

3. Multiples of 125: The number of multiples of 125 in 115! is given by ⌊115/125⌋ = 0 since there are no numbers in the range 1 to 115 that are multiples of 125.

Adding up these counts, we have 23 + 4 = 27 factors of 5.

Therefore, there will be 27 zeros at the end of 115!.

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Answer:

c

Step-by-step explanation:

3x3=9

4x3=12

Answer:

C

Step-by-step explanation:

3 cups of almonds / 4 cups of cashews

So to get 9 cups of almonds, we need to multiply almonds and cashews by 3

3*3 = 9 cups of almonds

4*3 = 12 cups of cashews

So the correct answer is C

The expression [tex]x^{2}[/tex] + 6x + 5 can be completed by transforming it into the form a(x - h)^2 + k.

To complete the square, we want to rewrite the quadratic expression x^2 + 6x + 5 in a perfect square trinomial form. We can achieve this by adding and subtracting a constant term inside the parentheses.

Starting with the given expression: x^2 + 6x + 5

To complete the square, we need to take half of the coefficient of x and square it. Half of 6 is 3, and squaring 3 gives us 9. So, we add and subtract 9 inside the parentheses:

x^2 + 6x + 5 = (x^2 + 6x + 9 - 9) + 5

Now, we can group the first three terms as a perfect square trinomial and simplify:

(x^2 + 6x + 9 - 9) + 5 = (x + 3)^2 - 9 + 5

Simplifying further, we have:

(x + 3)^2 - 4

Therefore, the expression x^2 + 6x + 5 can be written in the form a(x - h)^2 + k as (x + 3)^2 - 4.

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the linear approximation at z = 0 to sin(42) is A + Bz, where A = sin(42) and B is the coefficient of z, which is cos(42).

The linear approximation of a function f(x) at a point x = a is given by the equation f(x) ≈ f(a) + f'(a)(x - a). In this case, we want to approximate sin(42) at z = 0.

The derivative of the sine function is cos(x), so the derivative of sin(42) with respect to z is cos(42). Evaluating the derivative at z = 0, we have cos(42).

To find A in the linear approximation A + Bz, we substitute z = 0 into the original function sin(42) and obtain A = sin(42).

Therefore, the linear approximation at z = 0 to sin(42) is A + Bz, where A = sin(42) and B is the coefficient of z, which is cos(42).

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Answer:

Please upload the full question.....here the total number of outcomes is not mentioned and hence it can't be solved.

Answer:

Hence, the required probability of getting a number greater than 4, P(E) = 1/3.

Step-by-step explanation:

Answer:

y€ R : y greater than or equal to 1

C ≈ 62.83cm is the answer!

Hope this helps!

Answer: 52.7856cm

Step-by-step explanation:

Circumference of a circle = 2πr

Note that radius = Diameter / 2 = 16.8/2 = 8.4cm

Circumference = 2πr

= 2 × 3.142 × 8.4

= 52.7856cm

Therefore, the circumference of the circle is 52.7856cm

Answer:

just did it on edg, 2021

Hence, AB≈CD and BC ≈ DA

What is a parallelogram?

A four sided closed figure with all its sides parallel to its opposite side.

Consider two triangles ΔABD and ΔBCD

∠BCD=∠BCA( opposite angles are always equal)

∠ABD=∠BDC (AD||BC)

∠ADB=∠DBC(AD||BC)

ΔABD ≅ ΔBCD

AB≈CD by CPCTC

BC ≈ DA by CPCTC

Hence, proved

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The dimensions of the rectangle are length = 62.75 inches and width = 20.25 inches.

The given problem states that the length of a rectangle is two more than triple the width.

If the perimeter is 166 inches, what are the dimensions of the rectangle? Let's solve the problem,

Step 1

Given, The length of the rectangle = l

Width of the rectangle = w

The perimeter of the rectangle = 166 inches

The formula for the perimeter of a rectangle is,

Perimeter = 2(l + w)

So, 166 = 2(l + w)166/2 = l + w83 = l + w ----(1)

Step 2

According to the given problem, The length of a rectangle is two more than triple the width

Therefore,

l = 2 + 3w

Substitute this value in equation (1)

83 = (2 + 3w) + w

83 = 2 + 4w

83 - 2 = 4w

81 = 4w

w = 81/4

w = 20.25 (approx)

Step 3

We have width w = 20.25 inches.

We can find the length l by substituting w in l = 2 + 3w

So,

l = 2 + 3(20.25)

= 2 + 60.75

= 62.75

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Answer:

4x+3x+2x=180

9x=180

x=20

Answer:

132 dollars I think

Step-by-step explanation:

Answer:

The unbrella is 13.5 feet tall.

Step-by-step explanation:

35 / 28 = 1.25

1.25x = 16 7/8

x = (16 7/8) / 1.25

x = 13.5 ft

Answer:  1/5 of a jar

Step-by-step explanation:

1/2 times 2/5= .2 = 1/5.

Answer:

1/5 of a jar

Step-by-step explanation:

The first three terms of x[n] using the power series expansion are x[0] = 73, x[1] = 2, and x[2] = 13.

We can select the first three terms of x[n] using the power series development by expressing the given articulation X(z) as a polynomial in z. We should modify the articulation as follows to obtain the power series development: By comparing the given expression to the power series form, the coefficients can be identified: X(z) equals 2z3, 13z2, 7z, 73, 722/z, 2/z, and 1: a0 rises to 73, a1 approaches 2, a2 approaches 13, and a3 approaches 7. X(z) = a0, a1z, a2z2, a3*z3, and... Consequently, the following are the first three terms of x[n]:

The initial three terms of x[n] are provided by the power series development: x[0] = a0; x[1] = a1; x[2] = a2; x[0] = a0; The values of x[0] and x[1] are 73, 2, and 13.

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The z-score to test the hypothesis that the part is coming out longer than usual is approximately 1.61.

Sample mean = x = 12.20 cm

Population mean = μ = 12.05 cm

Standard deviation = σ = 0.350 cm

Sample size = n = 14

A hypothesis is an informed prediction regarding the solution to a scientific topic that is supported by sound reasoning. there is the expected result of the experimentation even if there is not proved in an experiment.

Calculating the z-score -

[tex]z = (x - u) / (\alpha / \sqrt n)[/tex]

Substituting the values -

[tex]z = (12.20 - 12.05) / (0.350 / \sqrt{14)[/tex]

= z = 0.15 / (0.350 / √14)

= 0.093

Substituting the value again into the formula:

z = 0.15 / 0.093

= 1.61

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Answer:

∠R = 38°

Step-by-step explanation:

∠R = 1/2(arc PE - arc SQ)

∠R = 1/2(140 - 64) = 1/2(76) = 38°

Answer:

77 meters

Step-by-step explanation:

For a right triangle, the formula for side length is

a^2 + b^2 = c^2 where c is the hypotenuse (opposite the right angle)

36^2 + b^2 = 85^2

1296 + b^2 = 7225

b^2 = 5,929

find the square root of both sides:

b = 77

Please let me know if you have questions.

Answer:

x = 5

x = 6

x = 10

Step-by-step explanation:

NOTE: IT HAS TO BE MORE THAN 5 OR EQUAL TO 5

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Answer:

Step-by-step explanation:

(-1,-9)

The partial derivatives are as follows :

(a) fx = 1 / (x + 2y^2 + 3z^3)

(b) fz = 3z^2 / (x + 2y^2 + 3z^3)

(c) d^2f/dzdx = -3z^2 / (x + 2y^2 + 3z^3)^2

To find the partial derivatives of the function f(x, y, z) = ln(x + 2y^2 + 3z^3), we differentiate with respect to each variable while treating the other variables as constants.

(a) Partial derivative with respect to x (fx):

To find fx, we differentiate the function f(x, y, z) with respect to x while treating y and z as constants. The derivative of ln(u) with respect to u is 1/u, so we have:

fx = d/dx ln(x + 2y^2 + 3z^3) = 1 / (x + 2y^2 + 3z^3)

(b) Partial derivative with respect to z (fz):

To find fz, we differentiate the function f(x, y, z) with respect to z while treating x and y as constants. Again, applying the derivative of ln(u), we get:

fz = d/dz ln(x + 2y^2 + 3z^3) = 3z^2 / (x + 2y^2 + 3z^3)

(c) Second partial derivative with respect to z and x (d^2f/dzdx):

To find d^2f/dzdx, we differentiate fz with respect to x while treating y and z as constants. We differentiate fx with respect to z while treating x and y as constants, and then take the derivative of the result with respect to z. It can be written as:

d^2f/dzdx = d/dx (d/dz ln(x + 2y^2 + 3z^3)) = d/dx (3z^2 / (x + 2y^2 + 3z^3))

= -3z^2 / (x + 2y^2 + 3z^3)^2

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Q1: Sum of first 68 positive odd integers.

Let's represent the first 68 positive odd integers by: 1, 3, 5, 7, ..., 135, 137. The first term, a = 1. The last term, l = 137And, the number of terms, n = 68We need to find the sum of these terms. To find the sum of an arithmetic series, we use the following formula: Sn = n/2[2a + (n-1)d]. Here, d = common difference. Since the given sequence is of odd numbers, the difference between any two consecutive terms is 2. So, d = 2. Put these values in the formula to get: Sn = 68/2[2(1) + (68-1)2], Sn = 34[2 + 135], Sn = 68 × 67Sum of first 68 positive odd integers = 4546.

Q2: Degree of recurrence relation. To find the degree of a recurrence relation, we find the largest value of n in the relation. Here, an = 2an-2 + 5an-49The largest value of n in the relation is n = 49. So, the degree of the recurrence relation is 49.

Q3: Number of ways to elect office bearers in an organization. Let's assume that the 19 members of the organization are named M1, M2, M3, ..., M19. The president can be elected in 19 ways. After the president is elected, the treasurer can be elected in 18 ways. After the treasurer is elected, the secretary can be elected in 17 ways. Therefore, the total number of ways in which the president, treasurer, and secretary can be elected is:19 × 18 × 17 = 5,814.

Q4: Greatest Common Divisor (GCD)To find the GCD of two numbers, we need to find their prime factors.28 × 37 × 58 = 2² × 7 × 37 × 2 × 29 = 2³ × 7 × 29 × 37Similarly, 23 × 33 × 54 = 23 × 3² × 2 × 3 × 3 × 2 × 3 = 2³ × 3⁵ × 23.

The common prime factors are 2³ and 23. So, the GCD is: 2³ × 23 = 184. The GCD is 184.

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The perimeter is the distance all the way around. So it's the sum of the lengths of all 4 sides.

From the picture, you can clearly see the lengths of all 4 sides.

Writum down and adum up !

Answer:

The amount earned for the week=$110

Amount earned from commission =$30

Step-by-step explanation:

commission earned on sales = $200×15%= $30

total amount for the week=$80 +$30= $110

The Taylor polynomial of degree 3 for the function 1/(1-2x) centered at x=0 is (1+2x+4x²+8x³).

Explanation: Given, 1/(1-2x) = ∑n=0 to infinity of 2^n * x^n The above series is the binomial series for (1+x)^n where n=-1Using the binomial series for n=-1, we have1/(1-2x) = ∑n=0 to infinity of 2^n * x^n= ∑n=1 to infinity of 2^(n-1) * x^(n-1)= 1 + ∑n=1 to infinity of 2^n * x^nTaking up to degree 3, we get1/(1-2x) = 1 + 2x + 4x² + 8x³ + ...Therefore, the Taylor polynomial of degree 3 for 1/(1-2x) is 1 + 2x + 4x² + 8x³.

An infinite sum of words that are expressed in terms of a function's derivatives at a single point is known as the Taylor series or Taylor expansion of a function in mathematics. Near this point, the function and the sum of its Taylor series are equivalent for the majority of common functions. for Brook Taylor, who introduced the Taylor series in 1715, they are named for him. In honour of Colin Maclaurin, who made great use of this unique example of Taylor series in the middle of the 18th century, a Taylor series is sometimes known as a Maclaurin series where 0 is the point at which the derivatives are taken into account.

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Answer:

17÷z = quotient

Step-by-step explanation:

quotient = ÷