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zimovet [89]
3 years ago
7

How many centimeters or in 400 Millimeters​

Mathematics
1 answer:
RUDIKE [14]3 years ago
8 0

There are 40 centimeters in 400 millimeters

Please rate my answer

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If measure FG = 120° and measure EH = 94, what is measure angle EPH?
tankabanditka [31]

Answer:

∠EPH = 107°

Step-by-step explanation:

∠EPH is an angle formed by two intersecting chords.

arc FG and arc EH are the arcs ∠EPH intercepts.

The relationship between an angle formed by intersecting chords and their intercepted arcs is that the angle is equal to half the sum of its intercepted arcs.

Which means that ∠EPH = 1/2 ( arc FG + arc EH )

If arc FG = 120° and arc EH = 94°

Then ∠EPH = 1/2 ( 120 + 94 )

120 + 94 = 214

214 / 2 = 107

Hence, ∠EPH = 107°

5 0
3 years ago
Read 2 more answers
What is the value of 1/3x-3/4 when x =1/4
satela [25.4K]

Answer:

The value of 1/3x-3/4 when x=1/4 is 0.08333 repeated.

Step-by-step explanation:

7 0
3 years ago
OMG last one please help
vichka [17]

Answer:

m<D = 14

Step-by-step explanation:

Exterior angles thm:

83 + 6x - 4 = 21x + 4

6x + 79 = 21x + 4

6x - 21x = 4 - 79

-15x = -75

x = 5

m<D:

6x - 4

6(3) - 4

18 - 4

14

3 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
F(x) = –3x² - 6x +1<br><br> Find f(8)
MAXImum [283]

Answer:

-239

Step-by-step explanation:

6 0
3 years ago
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