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kolezko [41]
3 years ago
7

( 3x + 5 ) ki power 2 expend it​

Mathematics
1 answer:
Pepsi [2]3 years ago
4 0

This question is soooo easy

Step-by-step explanation:

(3x)^2+(5)^2+2.3x.5

9x^2+25+30x

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Prove using coordinate geometry. Prove that the points (0,0), (4,0), (4, 4) and (0, 4) create a rhombus and then show that the m
BaLLatris [955]

Answer:

IDK

Step-by-step explanation:

IDK

5 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
What equation is parallel to 3y = 6x-10 and passes through A(4,5)?
elena-s [515]

Answer:

57?

Step-by-step explanation:

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2 years ago
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The amount of students that had more than 6 candy bars.

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3 years ago
Help! which one is it? I'm confused! :(
disa [49]

Answer:

its D

Step-by-step explanation:

x+2x=90-45

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