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asambeis [7]
3 years ago
15

Hurry i need help!!

Mathematics
1 answer:
diamong [38]3 years ago
4 0

Answer:

Take rabbit across

go back

Take fox across and go back across with rabbit

take carrots across and leave them with fox

go back and get rabbit

Step-by-step explanation:

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Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
1 year ago
Marcos purchased a sail boat for $56,900. He anticipates each year the boat will decrease in value by 8.1% from the previous yea
Varvara68 [4.7K]

Answer: 56,900\left( 0.919\right)^x

Step-by-step explanation:

Given

Marcos purchased a sailboat for \$56,900

Each year the boat will decrease in value by 8.1\%

After 1 year it is

\Rightarrow 56,900-56,900\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\\

after another year it becomes

\Rightarrow 56,900\left( 1-0.081\right)-56,900\left( 1-0.081\right)\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\cdot \left( 1-0.081\right)\\\Rightarrow 56,900\left( 1-0.081\right)^2

After x years it is

\Rightarrow 56,900\left( 1-0.081\right)^x

\Rightarrow 56,900\left( 0.919\right)^x

5 0
2 years ago
In the problem solving model, after you read the question, what do you do next?
KatRina [158]
This question mean no sense lol
3 0
2 years ago
A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school
Alla [95]

Answer:

The coach should start recruiting players with weight 269.55 pounds or more.                                                    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.15

P( X > x) = P( z > \displaystyle\frac{x - 225}{43})=0.15  

= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15  

=P( z \leq \displaystyle\frac{x - 225}{43})=0.85  

Calculation the value from standard normal z table, we have,  

P(z < 1.036) = 0.85

\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

3 0
3 years ago
Please help I’m so confused 25 points and brainliest!!
Soloha48 [4]

Answer:

This should be correct ( I havent done it in 3 years but it seems right)

Step-by-step explanation:

1.c≈15.71

2.c≈31.42

3.d=12.5

4 0
3 years ago
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