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2 years ago

The answer to this please and thank you

Mathematics

1 answer:

Tom [10]2 years ago

8 0

Answer:

Step-by-step explanation:

$4(7 {x}^{2} ) = 4 \times 7 {x}^{2} = 28 {x}^{2} \\$

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How many triangles are formed by the diagonals from one vertex of a regular octagon?

Natali [406]

6..............................................

6 0

3 years ago

If 60 marks is 40% , what is full marks

mixas84 [53]

So if 60 marks is 40%do buildind blocks 40%+ 40 % 80% = 120 marks so ad 20 % which is 30 marks so full marks is 150

6 0

2 years ago

Read 2 more answers

The cheetah is the fastest land animal, capiable of attaining a speed of 110 km/h. In August 2009 , the world record for the men

lozanna [386]

Answer:

About 3 times faster

Step-by-step explanation:

Cheetah's speed = 110 km / hr

World 100m sprint record :

Speed of sprinter :

Distance / time

100 m / 9.58s

= 10.438413 m/s

Convert Cheetah's speed to m/s

110 km/hr = (110 * 1000)m / (60 * 60 s)

= 110000 m / 3600 s

= 30.55555 m/s

Number of times cheetah is faster than sprinter :

30.55555 m/s ÷ 10.438413 m/s

= 2.927 times

5 0

2 years ago

A rope is 225 centimeters long.

n200080 [17]

Answer:

This is impossible, unless you meant centimeters. If you did, then the answer would be 113.

Step-by-step explanation:

225 - 112 = 113.

8 0

2 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

1 year ago