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Juli2301 [7.4K]
3 years ago
13

A water park charges $5 for entry into the park and an additional $2 for each of the big water slides. Steven spent $17 on his v

isit to the water park. Which equation can be used to find the number of big water slides, x, that Steven went on?
Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0
5+2x=17
That's what I got anyways. 
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Beacause 35 is less than 39 the frezzing point of butter is closer to the frezzing point of water which is 0
maria [59]

if 35 is LESS than 39

so +35 - 39 is -4 F°

the answer to the freezing point of (hehe) butter is -4 F°

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Match each equation to its solution.
astra-53 [7]
Good job you got it right! But others won’t be able to find it because you have a basic writing for it. Next time put the actual equations on so others can find it!
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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
Please help. I’ll mark you as brainliest if correct!
sesenic [268]

Answer:

Quantity (lbs) of type 1 candy  x = 8

Quantity (lbs) of type 2 candy  y = 17,5

Step-by-step explanation:

Let´s call  "x" quantity (in pounds) of candy type 1 in the mixture, and "y" quantity (in pounds ) of candy type 2, then according to the problem statement.

x  +  y  =  25,5

2,20*x  +  7,30*y = 5,70 * 25,5   ⇒  2,20*x  +  7,30*y = 145,35

Then we have a two equation system

x  +  y  =  25,5                                   ⇒   y = 25,5 - x

2,20*x  +  7,30*y = 145,35               ⇒ 2,20*x + 7,30* (25,5 - x ) = 145,35

2,20*x + 186,15 - 7,30*x = 145,35

5,1*x  = 40,8

x = 40,8/5,1

x = 8 lbs

And   y  =  25,5 - 8

y = 17,5 lbs

5 0
2 years ago
Complete your statement. Round to the nearest hundredth if necessary.
Vera_Pavlovna [14]

Answer:

?

Step-by-step explanation:

??????

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