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Kobotan [32]
3 years ago
11

(3 5/7+ 1 1/5)+ 2 1/7

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:7 2/35

Step-by-step explanation:

This is because of pemdas parenthesis exponents multiplication division addition and subtraction

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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
2 years ago
The sum of three consecutive integers is 63. What is the lowest of the three integers?
sveta [45]
  One number be x
x+(x+1)+(x+2)=63
x+x+1+x+2=63
3x+3=63
3x=63-3
3x=60
x=60:3
x=20
smallest number = x=20
middle <span>number = x+1= 20+1=21
</span>largest<span><span><span> number = x+2</span> =20+2= 22</span>
</span>
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2nd one behsbshsnnshs

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2 years ago
Citing drastic differences between her school and the white school across the
ella [17]

Answer:

the pay one

Step-by-step explanation:

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3 years ago
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The answer would be 26 because .1-4 rounds down, 5-9 rounds up
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