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-BARSIC- [3]
3 years ago
11

An example of consecutive odd integers is 23, 25, 27, and 29. Find four consecutive odd integers with a sum of 160. Show your wo

rk. After completing the problem, write out the steps, in words, that you used to solve the problem??????????????? Please tell me the steps to so I can understand please.
Mathematics
1 answer:
Andrew [12]3 years ago
8 0
Step1: Define an odd integer.
Define the first odd integer as (2n + 1), for n = 0,1,2, ...,
Note that n is an integer that takes values 0,1,2, and so no.

Step 2: Create four consecutive odd integers.
Multiplying n by 2 guarantees that 2n will be zero or an even number.
Therefore (2n + 1) is guaranteed to be an odd number.
By adding 2 to the odd integer (2n+1), the next number (2n+3) will also be an odd integer.

Let the four consecutive odd integers be
2n+1, 2n +3, 2n +5, 2n +7 

Step 3: require that the four consecutive integers sum to 160.
Because the sum of the four consecutive odd integers is 160, therefore
2n+1 + 2n+3 + 2n+5 + 2n +7 = 160
8n + 16 = 160
8n = 144
n = 18

Because 2n = 36, the four consecutive odd integers are 37, 39, 41, 43.

Answer: 37,39,41,43

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8

Step-by-step explanation:

you just count backwards from 5 until you reach -4

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3 years ago
Solve equation. 7/8-2/3
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Answer:

\frac{7}{8}  -  \frac{2}{3 } \\  \frac{7 \times 3}{8 \times 3}  -  \frac{2 \times 8}{3 \times 8 }  \\  \frac{21 - 16}{24 }  \\  =  \frac{5}{24}

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3 years ago
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To illustrate the effects of driving under the influence of alcohol, a police officer brought a DUI simulator to a local high sc
nexus9112 [7]

Answer:

Check the explanation

Step-by-step explanation:

Here we have to first of all carry out dependent sample t test. consequently wore goggles first was selected at random for the reason that the reaction time in an emergency taken with goggles would be greater than the amount of reaction time in an emergency taken with not so weakened vision. So that we will get the positive differences d = impaired - normal

b)

To find 95% confidence interval first we need to find sample mean and sample sd for difference d = impaired minus normal.

We can find it using excel that is in the first attached image below,

Therefore sample mean ( \bar{X}_{d} ) = 0.98

Sample sd ( \bar{S}_{d} ) = 0.3788

To find 95% Confidence interval we can use TI-84 calculator,

Press STAT ----> Scroll to TESTS ---- > Scroll down to 8: T Interval and hit enter.

Kindly check the attached image below.

Therefore we are 95% confident that mean difference in braking time with impaired vision and normal vision is between ( 0.6888 , 1.2712)

Conclusion : As both values in the interval are greater than 0 , mean difference impaired minus normal is not equal to 0

There is significant evidence that  there is a difference in braking time with impaired vision and normal vision at 95% confidence level .

7 0
3 years ago
The blades of a windmill turn on an axis that is 35 feet above the ground. The blades are 10 feet long and complete two rotation
a_sh-v [17]

The correct option is (c) h = 10sin(π15t)+35.

The equations can be used to model h, the height in feet of the end of one blade, as a function of time, t, in seconds is  h = 10sin(π15t)+35.

<h3>How do windmills rotate?</h3>

The blades of a turbine, which resemble propellers and function much like an airplane wing, capture the wind's energy.

A pocket of low-pressure air develops on one side of the blade when the wind blows. The blade is subsequently drawn toward the low-pressure air pocket, which turns the rotor.

Calculation for the equation of the model height-

Let's now review each choice individually and select the best one.

The blade is horizontal at time t = 0. As a result, h = 35 at t = 0 is valid for all of the possibilities in this situation.

They accomplish two spins in a minute. The blades will so complete one rotation in 30 seconds. and they will complete a quarter rotation in 15/2 seconds. Because of this, the blade will be vertically up from time t = 0 to t = 15/2. Its height in this instance should be 35 + 10 = 45 ft. Let's now examine the available possibilities.

If we put t=15/2 in the options

Option (a) gives h = 25

Option (b) gives h = -10sin(15/2) + 35

Option (c) gives h = 45

Option (d) gives h = 10sin(15/2) + 35

Therefore, the correct equation is given in option c.

To know more about windmill, here

brainly.com/question/2031508

#SPJ4

The complete question is -

The blades of a windmill turn on an axis that is 35 feet above the ground. The blades are 10 feet long and complete two rotations every minute. Which of the following equations can be used to model h, the height in feet of the end of one blade, as a function of time, t, in seconds? Assume that the blade is pointing to the right, parallel to the ground at t = 0 seconds, and that the windmill turns counterclockwise at a constant rate.

a) h = −10sin(π15t)+35  

b) h = −10sin(πt)+35

c) h = 10sin(π15t)+35

d) h = 10sin(πt)+35

6 0
2 years ago
Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1
aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

{3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\

3y^{2}.\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x}{y^{5}}

3 0
3 years ago
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