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3 years ago

Me when i was little i looked so weird

Mathematics

2 answers:

Lyrx [107]3 years ago

7 0

Answer:

Ok...

Step-by-step explanation:

givi [52]3 years ago

4 0

Once upon a time, there was a lovely princess. But she had an enchantment upon her of a fearful sort, which could only be broken by love's first kiss. She was locked away in a castle, guarded by a terrible fire-breathing dragon. Many brave knights had attempted to free her from the dreadful prison, but none prevailed. She waited in the dragon's keep, in the highest room of the tallest tower, for her true love and true love's first kiss." [Laughing] Like that's ever gonna happen.
[Paper Rustling, Toilet Flushes]
Shrek: What a load of--

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Solve the system by substitution 10x-2y=-16 and y=3x

lesantik [10]

The answer would be x=-4.

5 0

3 years ago

What is the arc measure, in degrees, of major arc \stackrel{\large{\frown}}{ADC} ADC ⌢ A, D, C, start superscript, \frown, end s

Mrrafil [7]

Answer:

$\stackrel{\large{\frown}}{ADC} = 186^{\circ}$

Step-by-step explanation:

Given

See attachment

Required

Determine the measure of $\stackrel{\large{\frown}}{ADC}$

$The\ sum\ of\ angles\ in\ a\ circle\ is$ $360^{\circ}$ .

So, we have:

$\stackrel{\large{\frown}}{ADC} + \stackrel{\large{\frown}}{APB} + \stackrel{\large{\frown}}{BPC} = 360^{\circ}$

Where:

$\stackrel{\large{\frown}}{APB} = 70^{\circ}$

$\stackrel{\large{\frown}}{BPC} = 104^{\circ}$

Substitute these values in the above equation.

$\stackrel{\large{\frown}}{ADC} + 70^{\circ} +104^{\circ} = 360^{\circ}$

$\stackrel{\large{\frown}}{ADC} + 174^{\circ} = 360^{\circ}$

Collect Like Terms:

$\stackrel{\large{\frown}}{ADC} = 360^{\circ} - 174^{\circ}$

$\stackrel{\large{\frown}}{ADC} = 186^{\circ}$

7 0

3 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

Where is point b on the number line?

Yuliya22 [10]

Answer:

i dont know but can you pick me brainiest

Step-by-step explanation:

4 0

3 years ago

Does anyone know how many ways you can make 16 and 19 on a grid

sasho [114]

You can look for the number 16 on the x-axis. (bottom line) and put a finger there. Then, look for 19 on the y -axis (left line) and put a finger there. Trace your left finger to the right and your right hand up until they meet. That is (16,19)

8 0

3 years ago