the price of fruit acid 1.65 for 11 ounces. Fruit B costs 2 cents more per ounce. What is the cost of 16 ounces of fruit B?

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Answer 1
Answer:

Answer: 2.72 i guess


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The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

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The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

The answer to this question is 0.002

203.530 divided by 0.7=

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8 is to 32 as 9 is to N? What is N?

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We have given 8 is to 32 as 9 is to N. After figuring that 4 is multiple,  So, the value of N is 36.

What is the unitary method?

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

We have given 8 is to 32 as 9 is to N.

We need to find the value of N.

8 times 4 is 32.

After figuring that 4 is multiple,

We can see that multiplying 9 by 4 gives the N.

9 times 4 is 36.

So, the value of N is 36.

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36, 8 times 4 is 32. After figuring that 4 is the multiple, you would multiply 9 by 4 giving you the N.

Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9) The amount of work required is ____ (Simplify your answer.)

Answers

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

\bf \int_(C)F

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write  

C = \bf C_1+C_2

where  

\bf C_1 =  line segment from (1, 5) to (0, 0)  

\bf C_2 = line segment from (0, 0) to (0, 9)

so,

\bf \int_(C)F=\int_(C_1)F+\int_(C_2)F

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1

Hence \bf C_1 can be parameterized as

\bf r_1(t) = (1-t, 5-5t) for 0 ≤ t ≤ 1

and \bf C_2 can be parameterized as

\bf r_2(t) = (0, 9t) for 0 ≤ t ≤ 1

The derivatives are

\bf r_1'(t) = (-1, -5)

\bf r_2'(t) = (0, 9)

and

\bf \int_(C_1)F=\int_(0)^(1)F(r_1(t))\circ r_1'(t)dt=\int_(0)^(1)(5-5t,t-1)\circ (-1,-5)dt=0

\bf \int_(C_2)F=\int_(0)^(1)F(r_2(t))\circ r_2'(t)dt=\int_(0)^(1)(9t,0)\circ (0,-9)dt=0

In consequence,

\bf \int_(C)F=0

Final answer:

The work done by the force field F = (y, -x) along the given path is -5 Joules. This was calculated by breaking the path into two segments and calculating the work done for each segment.

Explanation:

To calculate the work done by the force field F = (y, -x) when moving an object along a specific path, we utilize the concept of the line integral or the dot product of the force and the displacement vector. We can break down the given path into two line segments and solve each separately.

The first segment is from (1, 5) to (0, 0). Only the x component of the displacement is non-zero here, the force is F = (5, -1). Thus the work done on this segment is given by W = F.d = Fd cos θ = -(5 N)(1 m)(cos(180)) = -5 J, where J stands for Joules, the unit of work or energy.

The second segment is from (0, 0) to (0, 9). The force and displacement are perpendicular so the work done is 0.

By adding the work done on these two segments, we arrive at the total work done: W_total = -5 J + 0 J = -5 J

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Determine the axis of
symmetry for y=x²+2

Answers

Answer:

x=0

Step-by-step explanation:

The axis of symmetry always passes through the vertex of the parabola . The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For a quadratic function in standard form, y=ax2+bx+c , the axis of symmetry is a vertical line x=−b/2a

in our case

y=x^2+2

a=1 b=0 c=2

so x=-b/2a=-0/2*1=0

x=0 is axis of symmetry

A community pool office two types of memberships monthly and annual memberships. At the beginning of the year the ratio of monthly annual memberships is 10 to 3. However the pool offered an incentive to have members move the annual membership. After the incentive the ratio is 5 to 8. If there are 50 monthly members after the incentive and how many monthly members were there before?

Answers

Answer:

The number of monthly memberships before the incentive was 100

Step-by-step explanation:

Remember that

The total memberships after the incentive is equal to the total memberships before the incentive

step 1

Find out the annual memberships after the incentive

Let

x -----> monthly memberships after the incentive

y -----> annual memberships after the incentive

we know that

-----> equation A

substitute the value of x in equation A

step 2

Find out the total memberships after the incentive

step 3

Find out the monthly members before the incentive

Let

x -----> monthly memberships before the incentive

y -----> annual memberships before the incentive

we know that

-----> equation A

-----> equation B

substitute equation A in equation B and solve for x

therefore

The number of monthly memberships before the incentive was 100

Final answer:

Initially, there were 100 monthly members at the community pool, before the incentives were offered.

Explanation:

We start with the information that the ratio of monthly to annual memberships was initially 10 to 3, and then became 5 to 8. After the incentive, we're told there are now 50 monthly members.

To solve this problem, we set up a proportion. Since each part of the new ratio equals 10 members (50 monthly members/5 parts = 10 members per part), we can infer that before the incentive there were 10*10=100 monthly members.

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