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Strike441 [17]
3 years ago
9

A colony of Insects doubles every day. If the colony has 5 insects today, How

Mathematics
1 answer:
Alla [95]3 years ago
5 0

Answer: 2621440

Step-by-step explanation: just multiply by 2 every time

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Can someone please help me answer these 2 questions with a full explanation of how you got the answer so I can do the rest mysel
finlep [7]

(a) The measure of the indicated angle for figure 1 is 48⁰.

(b) The measure of the indicated angle for figure 2 is 42⁰.

<h3>Measure of the indicated angle</h3>

The measure of the indicated angles can be calculated as follows;

<h3>Figure 1</h3>

sinθ = opp/hypo

sinθ = 72/97

sinθ = 0.7423

θ = sin⁻¹(0.7423)

θ = 47.9⁰ ≈ 48⁰

<h3>Figure 2</h3>

sinθ = 65/97

sinθ = 0.6701

θ = sin⁻¹(0.6701)

θ = 42.1⁰ ≈ 42⁰

Learn more about angles here: brainly.com/question/25770607

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7 0
2 years ago
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At an outdoor party location, the tables seat 12 guests each. The party host invited 163 people and 9 people responded that they
dalvyx [7]

Answer:

ok

Step-by-step explanation:

8 0
3 years ago
Please help ASAP thank you
vladimir1956 [14]

Answer:

Triangle.

Step-by-step explanation:

3 0
3 years ago
What is a bank statement
azamat

Answer:

A bank statement can be defined as a record of payments in and out of a bank account.

Step-by-step explanation:

For example, energy providers offer direct debit payments as the amount that is used each month may vary and they take the required amount at each bill. Standing orders are similar to a direct debit except that they are for a fixed amount.

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
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