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Olenka [21]
3 years ago
14

3+m+4≤5 please help due tommorow

Mathematics
2 answers:
ruslelena [56]3 years ago
7 0
3+m+4 \leq 5 \\\\ m+7 \leq 5 \\\\ m \leq 5-7 \\\\ m \leq -2 \\\\ \boxed{m=(-\infty;-2]}
algol [13]3 years ago
7 0
3+m+4≤5
m+7≤5
m+7-7≤5-7 (you have to subtract 7 on both sides)
m≤-2
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Can u help me with my work don’t send a link !!!
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Answer:

A and C

Step-by-step explanation:

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6 0
3 years ago
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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 is 32.5 billion in scientific notation
Nana76 [90]
It would be 3.25 x 10¹⁰
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3 years ago
H(t)=-10t+6<br> H(_____)=-44<br> Help me please.
alexira [117]

Answer:

-10t +6=-44

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8 0
3 years ago
A large emerald with a mass of 378.24 grams was recently discovered in a mine. If the density of the emerald is 2.76grams over c
bogdanovich [222]

Answer:

Volume of emerald = 137.04cm³

Step-by-step explanation:

Given that:

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The equation for this relationship is ρ = m / V in which ρ is density, m is mass (kg or g) and V is volume (cm³ or m³) making the density unit kg/m³ or g/cm³.

Therefore

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Volume of emerald = Mass of emerald / Density of emerald

Volume of emerald = 378.24grams / 2.76grams/cm³

Volume of emerald = 137.043cm³ ~137.04cm³

Volume of emerald =137.04cm³

6 0
3 years ago
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