b. Can an objective function P=a x+b y+c have (the same) maximum value at all four vertex points Q, R, S , and T ? At points R and S only? Explain using examples.

The account in which there was a mistake in the calculation of the simple interest is account B.

We know that the interest is an amount of money that is obtained on top of the initial money that was deposited. The initial money that you deposit in the account is called the principal. The rate is the percentage at which the interest is calculated and the time is the period that has elapsed since the deposit was made.

Given that;

I = PRT/100

I = interest

P = principal

R = rate

T = time

For account A

I = 200 * 3.9 * 1/100

I = $7.8

For account B;

I = 400 * 2.7 * 1/100

I = $10.8

We can see that there must have been a calculation error as the interest in account B was being calculated.

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Students can plant seeds in whole community garden in 5 1/3 h. Hence, Vanessa is not correct.

In the given question, it takes 2 h for students to plant seeds in 3/8 of a community garden.

Vanessa says it will take 4 h for the students to plant seeds for the whole garden.

We have to check Vanessa is correct or not.

Students plant seeds in 3/8 of the community garden in = 2 h

So, Students can plant seeds in 1 of the community garden in = 2/(3/8) h

Students can plant seeds in 1 of the community garden in = 2/3 *8 h

Students can plant seeds in 1 of the community garden in = 16/3 h

Students can plant seeds in 1 of the community garden in = 5 1/3 h

Hence, students can plant seeds in whole community garden in 5 1/3 h. Hence, Vanessa is not correct.

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Answer:

$740

Step-by-step explanation:

(see attached picture for reference)

I'm going to assume that when you wrote "500", you mean "$500"

recall that the formula for compound interest:

A=P(1+r/n) ^ (nt)

where

P = principal amount = $500

r = interest rate = 8% = 0.08

n = number of times compounded in a year = 2 (note: semi-annually means twice a year, hence n = 2)

t = number of years = 5

substituting this into the equation,

A=P(1+r/n) ^ (nt)

A=500 [ 1+(0.08/2) ] ^ (2 x 5)

A= $740.1221

A = $740 (rounded to nearest dollar)

The given equation is expressed as

s(t) = t^3 - 5t^2 + 3t

This is a position function. To find the velocity function, we would differentiate the given function with respect to t. It becomes

v(t) = 3t^2 - 10t + 3

To find the time when the velocity is zero, we would substitute v(t) = 0 into the equation, It becomes

3t^2 - 10t + 3 = 0

This is a quadratic equation. We would solve for t by applying the method of factorisation. The first step is to multiply 3t^2 with 3. It becomes 9t^2. We would find two terms such that their sum or difference is - 10t and their product is 9t^2. The terms are - t and - 9t. By replacing -10t with - t - 9t, we have

3t^2 - t - 9t + 3 = 0

By factorising by grouping, we have

t(3t - 1) - 3(3t - 1) = 0

Since 3t - 1 is common, it becomes

(3t - 1)(t - 3) = 0

3t - 1 = 0 or t - 3 = 0

3t = 1 or t = 3

t = 1/3 or t = 3

The particle is at rest at t = 1/3 and t = 3

Answer:

100% = 30. There are 30 students in Mr.Fulton's class

Step-by-step explanation:

If we have 12 as 40%, we divide it by 2 to get the base value which is 6=20%.

Now we multiply the base by 5 so 20% x 5= 100% and 6 x 5 equals 30. so 100% = 30

Answer:

there are 30 students

Step-by-step explanation:

Answer:

see below

Step-by-step explanation:

The second equation tells us x = y - 3 so when we plug in y - 3 for x in the first equation we get 4(y - 3) + 3y = -5 → 4y - 12 + 3y = -5 → 7y = 7 → y = 1. Plugging this value into the second equation we get x = y - 3 = 1 - 3 = -2.

Answer:

(-2,1)

Step-by-step explanation:

4x+3y=-5

x=y-3

first, plug in  y-3 for x to find y:

4(y-3)+3y=-5

4y-12+3y=-5

7y-12=-5

7y=7

y=1

now, plug in 1  for y to find x:

4x+3*1=-5

4x+3=-5

4x=-8

x=-2

(-2,1)

Answer:

336

Step-by-step explanation:

The exclamation mark in math represents the factorial operation, which means you multiply the number by decreasing positive integers.

Therefore:

5! = 5 x 4 x 3 x 2 x 1 = 120.

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

8!/5! = 40320/120

40320/120 = 336

The index variable for the element at the last row and last column in the array x is B. x[4][6].

In Java, arrays are zero-indexed, which means the first element of an array is at index 0. The given array x is a 2D boolean array with dimensions 5x7 (5 rows and 7 columns). To access the last row and last column, we need to subtract 1 from each dimension:

- For the last row, the index is 5 - 1 = 4.
- For the last column, the index is 7 - 1 = 6.

So, the correct index variable to access the last row and last column is x[4][6]. The other options provided are incorrect as they do not accurately represent the index of the last element:

- a. x[4][5] refers to the second to last element in the last row.
- c. x[5][6] would be out of bounds, as the maximum row index is 4.
- d. x[5][7] would also be out of bounds, as both the row and column indices are too large.

Therefore, the correct option is B.

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Answer: $2100

Step-by-step explanation:

Since 12.5 percent of 2400 is 300, Just subtract 300 from 2400 to get the selling price with no sales tax. And the final answer is $2100.

The power rule for logarithms states that logb(M^p) = p * logb(M). The logarithm of a number with an exponent is the product of the exponent and the logarithm of that number.

The power rule states that logb(M^p) = p * logb(M), where M, b, and p are positive real numbers.

To understand this rule, let's consider an example. Suppose we have log2(8^2). According to the power rule, this is equivalent to 2 * log2(8).

In this case, M is 8, b is 2, and p is 2. The power rule tells us that we can bring the exponent (p) down and multiply it with the logarithm of the base (b) raised to the number (M).

So, log2(8^2) can be simplified as 2 * log2(8), which is 2 * 3 = 6.

In general, the power rule allows us to simplify logarithmic expressions by bringing the exponent down and multiplying it with the logarithm of the base.

This rule is particularly useful when dealing with complex logarithmic expressions and simplifying them to a more manageable form.

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The equation that can be used to find the number of hours required to make a certain number of leashes is L = (15/2) * h.

Let's denote the number of hours as "h."

Based on the given information, we know that Bob can make 7 leashes in 2 hours, which can be expressed as a ratio:

Bob's rate: 7 leashes / 2 hours = 7/2 leashes per hour

Similarly, Nina can make 4 leashes in 1 hour, which can be expressed as a ratio:

Nina's rate: 4 leashes / 1 hour = 4/1 leashes per hour

To find the combined rate of Bob and Nina, we can add their individual rates:

Combined rate: (7/2 + 4/1) leashes per hour

To find the number of hours required to make a certain number of leashes, we can set up an equation using the combined rate:

Number of leashes = Combined rate * Number of hours

Let's assume we want to find the number of hours required to make "L" leashes. The equation becomes:

L = (7/2 + 4/1) * h

Simplifying the equation:

L = (7/2 + 8/2) * h

L = 15/2 * h

Therefore, the equation that can be used to find the number of hours required to make a certain number of leashes is L = (15/2) * h.

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Events Q and R are not mutually exclusive, thus we are given two probabilities, P(Q) = 0.41 and P(R) = 0.44.

Therefore,

Events Q and R are not mutually exclusive, thus we are given two probabilities, P(Q) = 0.41 and P(R) = 0.44.. In this instance, the likelihood of Q and R is equal to P(Q) × P(R), which is 0.1804. The solution to this issue is this.

The complete question is:

Suppose Q and R are independent events. Find the probability of Q and R when P(Q)=0.41 Ans P(R)=0.44.

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The estimated probability that 3 or more kicks are made out of 5 attempts, with a 72% probability of success per kick, is approximately 82.44%.

To estimate the probability that 3 or more kicks are made, we can use the binomial distribution.

Let X be the number of successful kicks in 5 attempts. Then X follows a binomial distribution with parameters n = 5 and p = 0.72.

The probability of making exactly k kicks is given by the binomial probability mass function

\(P(X=k) = (n choose k) * p^k * (1-p)^{n-k}\)

where (n choose k) is the binomial coefficient, which gives the number of ways to choose k successes out of n trials.

To find the probability of making 3 or more kicks, we can sum the probabilities of making 3, 4, or 5 kicks

P(X >= 3) = P(X=3) + P(X=4) + P(X=5)

\(P(X=k) = (n choose k) * p^k * (1-p)^{n-k}\)

P(X=3) = (5 choose 3) * 0.72³ * (1-0.72)⁵⁻³ = 0.2787

P(X=4) = (5 choose 4) * 0.72⁴ * (1-0.72)⁵⁻⁴ = 0.3769

P(X=5) = (5 choose 5) * 0.72⁵ * (1-0.72)⁵⁻⁵ = 0.1688

P(X >= 3) = 0.2787 + 0.3769 + 0.1688 = 0.8244

Therefore, the estimated probability that 3 or more kicks are made is approximately 0.8244, or 82.44%.

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Answer:

600 cm³

Step-by-step explanation:

12 x 10 = 120

120 x 5 = 600

Answer:

The middle one

Step-by-step explanation:

Answer: The Middle Graph - represents a nonlinear function

Answer:

Yes

Step-by-step explanation:

In Isosceles triangle, any 2 angles must be equal.

The angle measurements are 68°, 68°, and 44°.

Here,

two angles are equal, (i.e) 68°.

Hence,

the given angles can form an isosceles triangle.

Answer:

It's isosceles since two angles are equal

the solution for the initial value problem is: u(x, t) = sin(sqrt(-λ² * (a / 4)) * x) * exp(-λ² * t) where λ = ± sqrt(-4n² / a), and n is a non-zero integer.

The given partial differential equation is:

au(x, t) = 4 * (d²u(x, t) / dt²) / (dx²)

a) BCs (Boundary Conditions):

We have u(0, t) = 0 and u(π, t) = 0.

IC (Initial Condition):

We have u(x, 0) = x.

To solve this initial value problem, we need to find a function f(x) that satisfies the given boundary conditions and initial condition.

The solution for f(x) can be found using the method of separation of variables. Assuming u(x, t) = X(x) * T(t), we can rewrite the equation as:

X(x) * T'(t) = 4 * X''(x) * T(t) / a

Dividing both sides by X(x) * T(t) gives:

T'(t) / T(t) = 4 * X''(x) / (a * X(x))

Since the left side only depends on t and the right side only depends on x, both sides must be equal to a constant value, which we'll call -λ².

T'(t) / T(t) = -λ²

X''(x) / X(x) = -λ² * (a / 4)

Solving the first equation gives T(t) = C1 * exp(-λ² * t), where C1 is a constant.

Solving the second equation gives X(x) = C2 * sin(sqrt(-λ² * (a / 4)) * x) + C3 * cos(sqrt(-λ² * (a / 4)) * x), where C2 and C3 are constants.

Now, applying the boundary conditions:

1) u(0, t) = 0:

Plugging in x = 0 into the solution X(x) gives C3 * cos(0) = 0, which implies C3 = 0.

2) u(π, t) = 0:

Plugging in x = π into the solution X(x) gives C2 * sin(sqrt(-λ² * (a / 4)) * π) = 0. To satisfy this condition, we need the sine term to be zero, which means sqrt(-λ² * (a / 4)) * π = n * π, where n is an integer. Solving for λ, we get λ = ± sqrt(-4n² / a), where n is a non-zero integer.

Now, let's find the expression for u(x, t) using the initial condition:

u(x, 0) = X(x) * T(0) = x

Plugging in t = 0 and X(x) = C2 * sin(sqrt(-λ² * (a / 4)) * x) into the equation above, we get:

C2 * sin(sqrt(-λ² * (a / 4)) * x) * C1 = x

This implies C2 * C1 = 1, so we can choose C1 = 1 and C2 = 1.

Therefore, the solution for the initial value problem is:

u(x, t) = sin(sqrt(-λ² * (a / 4)) * x) * exp(-λ² * t)

where λ = ± sqrt(-4n² / a), and n is a non-zero integer.

Note: Please double-check the provided equation and ensure the values of a and the given boundary conditions are correctly represented in the equation.

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Answer:

Step-by-step explanation:

and 2.3333333333

Answer:

14x cannot be simplified any further.

Answer:

No, 14x can not be further simplified.

Step-by-step explanation:

I hope this helps.

The combination formula helps to determine the number of non-negative solutions of equation

x₁ + x₂ + x₃ + x₄ + x₅ = 21 in different cases.

a) Total number of solutions for equation in case of x₁ ≥ 1, is equals to the 10626.

b) Total number of solutions for equation in case of xᵢ ≥ 2 for i = 1, 2, 3, 4, 5, is equals to the 1365.

c) Total number of solutions for equation in case of 0 ≤ x₁ ≤ 10 is equals to the 11649.

d) Total number of solutions for equation in case of 0 ≤ x₁ ≤ 3, 1 ≤ x₂ < 4, and x₃ ≥ 15, is equals to the 76.

We have an equation x₁ + x₂ + x₃ + x₄ + x₅ = 21, where xᵢ , i = 1, 2, 3, 4, 5 and we have to determine the number of solution of equation.

a) Solution of equation when x₁ ≥ 1

Let there are 5 boxes namely, x₁,x₂,x₃,x₄,x₅ in which 21 balls are placed distinguished. First ball is placed in box x₁. So, remained 20 balls are placed into 5 boxes with repititions. So, number of ways are C(n+r−1, r-1) = (n + r − 1)!/(r-1)!n! , where n = 20 , r = 5

C(n + r −1, r- 1) = C(20 + 5− 1 ,4)

= C(24,4) = 10626

b) Solution of equation for xᵢ ≥ 2 for i = 1,2,3,4,5.

Now, xᵢ ≥ 2 for i = 1,2,3,4,5, means first drop two balls into every box. Then, only 11 balls, remaining

n =11, r = 5

C(n+r−1, r-1) = (n+r−1)!/(r-1)!n!

= C(11 + 5−1, 4) = C(15, 4)

= 1365

c) Solution of equation for 0 ≤ x1 ≤ 10

when 21 balls dropped into five boxes x₁, x₂, x₃, x₄,x₅ then C(n+r−1, r-1) = (n+r−1)!/(r-1)!n! ; n= 21, r = 5

C(n + r −1, r - 1) = C(21+ 5− 1 ,4)

C(25, 4) = 12650

xi ≥ 11 , n=11, r = 5 then

C(n + r−1, r- 1 ) = C(11+4−1,4)

C(14,4) = 1001

C(25,4) − C(14,4) = 12650 − 1001 = 11649

d) Solution of equation for 0 ≤ x₁≤ 3, 1 ≤ x₂ < 4, and x₃≥ 15. For this condition solution is = { x1≥ 0, x2≥ 0, x3 ≥ 15} - { x₁≥3, x₃≥ 15} - { x₂≥4, x₃≥ 15} - { x₁=0, x₃ ≥ 15}

Now, first 15 balls are dropped into x3 and remaining 6 are droped into remaining boxes, { x₁ ≥ 0, x₂≥ 0, x₃≥ 15}, Then, n = 6, r = 5

C(n + r −1, r - 1) = C(6+ 5− 1 ,4)

C(10, 4) = 210

Also, { x₁ ≥3, x₃ ≥ 15}, represents 3 and 15 balls are dropped into x₁ and x₃ respectively. and remaining 3 balls will dropped into 5 boxes with repititions, n = 3, r = 5

C(n + r −1, r - 1) = C(3+ 5− 1 ,4)

C(7, 4) = 35 ways

{ x₂≥4, x₃≥ 15} , represents 4 and 15 balls are dropped into x₂ and x₃ respectively. and remaining 2 balls will dropped into 5 boxes with repititions, n= 2, r = 5

C(n + r −1, r - 1) = C(5+ 4− 1 ,4)

C(7, 3) = 15 ways

{ x₂ = 0, x₃ ≥ 15}, represents 0 and 15 balls are dropped into x₂ and x₃ respectively. and remaining 6 balls will dropped into 4 boxes with repititions, n= 6, r= 4

C(n + r −1, r - 1) = C(6+ 4− 1 ,3)

C(9, 3) = 84 ways

So, the number of solutions for this case = 210 - 35 - 15 - 84 = 76.

Hence there are 76 non-negative integer solutions.

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Answer:

Step-by-step explanation:

the rectangle's area is simply: A = 4 x 8 = 32

the circle's area is : π x 2 x 2 so A = 4π

the half circle's area is : 4π / 2 = 2π

And the final area is A = 32 - 2π = 2 x (16-π)

In rectangle

Area:-

In semicircle

Area:-

Total area=

Answer:

0.9545

Step-by-step explanation:

Let an integer be x.

If x is increased by 15%, then the value of the number would be

115/100x=1.15x

If 1.15x is decreased by 17%, then the value of the number would be:

(100-17)/100×1.15x

83/100×1.15x=0.9545x

Answer:

0.9545  

Step-by-step explanation:

15 % increase

17% decrease

increase followed by decrease

Answer:

\(\boxed{x = 7~~and~~y = -3}\)

.

Step-by-step explanation:

5x + 3y = 26 (times 6) 30x + 18y = 156

6x + 7y = 21 (times 5) 30x + 35y = 105

.

30x + 18y = 156

30x + 35y = 105    

-17y = 51

\(y = \frac{51}{-17}\)

y = -3

.

Substitute the value of y to one of equations

6x + 7y = 21

6x + 7(-3) = 21

6x - 21 = 21

6x = 21 + 21

6x = 42

\(x = \frac{42}{6}\)

x = 7

To make a 1.5 mg/ml solution from a 10 mg/ml stock solution, 750 µl of the stock must be taken.

To create a 1.5 mg/ml solution from a 10 mg/ml stock solution, we need to use the formula C1V1=C2V2. C1, the concentration of the stock solution, is 10 mg/ml. V1 is the volume of the stock solution, and C2 is the desired concentration, 1.5 mg/ml. V2 is the final volume, 5 ml. We can solve for V1 by rearranging the equation to V1 = (C2V2)/C1. Therefore, V1 = (1.5 x 5)/10 = 0.75 ml = 750 µl. We need to take 750 µl of the 10 mg/ml stock solution to make a 1.5 mg/ml solution with a total volume of 5 ml. To make a 1.5 mg/ml solution from a 10 mg/ml stock solution, 750 µl of the stock must be taken.

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Answer:

A=3, B=4, C=-5

Step-by-step explanation:

So A represents the amplitude, which is the distance between the midline (C), the middle of the graph, and the maximum (or minimum). B is the distance of the period of the graph. In this case, we see that the midline is C=-5 because that is the halfway point of the sine function. We also see that the period distance is B=5-1=4, so our period would be 2π/4 or π/2. Our amplitude would be A=3 because |a|=|-5-(-2)|=|-5+2|=|-3|=3. Observe the graph to see this visually.

Step-by-step explanation:

Let's start by assigning a variable to Kate's current age. We'll call it "K".

According to the problem, Kate's mother is 32 years older than Kate, so her mother's current age can be expressed as "K + 32".

In 4 years' time, Kate will be "K + 4" years old, and her mother will be "K + 32 + 4 = K + 36" years old.

The problem also tells us that in 4 years' time, Kate's mother will be triple Kate's age. We can express this information as an equation:

K + 36 = 3(K + 4)

Now we can solve for K:

K + 36 = 3K + 12

2K = 24

K = 12

Therefore, Kate is currently 12 years old.

Answer:

Kate is 12

Step-by-step explanation:

First add 4 to the mom's age then divide it by 3

Answer: Keep the compass needle on A, and stretch the compass width to B

Step-by-step explanation: If AB is also a line segment, then the answer is: Keep the compass needle on A, and stretch the compass width to B.

Using the concept of unit rate we calculate the cost of mowing 6 laws as $150 .

Rates and ratios frequently vary with regard to time, geography, a particular object (or group of objects), etc. They are typically mathematical functions as a result.

The charge to mow one lawn is $25 . This is the unite rate of the lawn.

Therefore the charge for 6 lawns is  = $25 × 6 = $150

Hence the cost to mow 5 lawns is $150.

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