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Anastaziya [24]
3 years ago
9

Cual es el valor absoluto de 26​

Mathematics
1 answer:
tia_tia [17]3 years ago
3 0
El valor absoluto de 26 is 26 |26|=26
Puedes decir que el valor absoluto de un número es el mismo número solo tiene que ser positivo
|-26|=26
|2|=2
|-59|=59
|-1/2|=1/2
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Please help!! I will give brainliest and no links!! Also please do it in a picture model “standard algorithm” please do both que
Alexandra [31]

Answer:

1. 219.571428571

2. 150.314285714

5 0
3 years ago
How do I simply this, thanks!
stepan [7]
(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4
7 0
3 years ago
Someone help me with this ASAP
Archy [21]

Answer:

1:6

3/5×2=6/10

2:14

21÷3=7

2×7=14

8 0
2 years ago
Read 2 more answers
You want to buy a $219,000 home. You plan to pay 10% as a down payment, and take out a 30 year fixed loan for the rest.
vovangra [49]

Answer:

see explanation

Step-by-step explanation:

I assume you need to find the 10% and also the amount per year maybe?

so, you will pay 21,900 dollars upfront

219000-21900=197100

219000*1/10=21900, which is ten percent.

197100=30x, x being money paid per year in this case

divide both sides by 30

6570=x

You will pay 6,570 each year for this house. You can continue to divide to find monthly payments. or quarterly payments. Divide by 12 to find monthly payment and  divide by 4 to find quarterly

5 0
3 years ago
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
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