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3 years ago

Answer for 49 Points

Mathematics

2 answers:

valentinak56 [21]3 years ago

4 0

Kayla has 2/14 of her allowance left. This is because she spent 12/14 of her allowance.

I hope this helps.

damaskus [11]3 years ago

4 0

the answer is 49 and 49 only

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Help me please with this question

docker41 [41]

Answer:

P=152

A=79

Step-by-step explanation:

add all the sides

6 0

3 years ago

A square has a perimeter of 39.6 millimeters. What is the length of each side?

garri49 [273]

Answer:

9.9 millimeters

Step-by-step explanation:

To find the perimeter of a square,

Equation: 4 • length = P

So in this case, we're just working backwards. To find the length, divide.

39.6/4 = 9.9

3 0

2 years ago

Read 2 more answers

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

Can all quadrilaterals be inscribed in a circle? Explain

-BARSIC- [3]

Depends of the size of the quadrilateral in relation to the circle, but otherwise, yes

3 0

3 years ago

At what elevation does the atomosphere become too thin to breath

prohojiy [21]

At around 20'000ft. and higher.

4 0

3 years ago