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riadik2000 [5.3K]
3 years ago
12

One winter day, the temperature increased from a low -5 degrees F to a high of 40 degrees F.BY how many degrees did the temperat

ure change ?
Mathematics
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

The temperature changed by 45˚

Step-by-step explanation:

Set up an algebraic equation

-5 + x = 40

Add 5 to both sides

x = 45

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Maria drove from Los Angeles (elevation 330 feet) to Death Valley (elevation –282 feet). What is the difference in elevation bet
motikmotik

Answer:

612 feet

Step-by-step explanation:

LA is located at 330 feet ABOVE SEA LEVEL

Death Valley is located 282 feet BELOW SEA LEVEL

We let the sea level be at 0 (consider a number line).

So,

LA would be at +330 feet

and

Death Valley would be at -282 feet

The elevation change between the two would be the difference:

330 - (-282) = 330 + 282 = 612 feet

The difference in elevation = 612 feet

3 0
2 years ago
How many tbsp of honey is needed for 20 cups of flour?
mr_godi [17]

Answer:

16 TBSP

Step-by-step explanation:

I did the math on a calculator thought I owed it to you, took me forever.

; ) plz mark brainliest.

5 0
2 years ago
Someone please HELPP me
Mazyrski [523]

Answer:

32 remaining cups for Kin

Step-by-step explanation:

7 0
2 years ago
Question 1
blondinia [14]
50 percent because 10 cars would be 100 percent so 5 cars would be 50 percent you just have to find the half
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
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