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STatiana [176]
2 years ago
5

Given that G=ab find the percentage increase in G when both a and b increase by 10%

Mathematics
1 answer:
monitta2 years ago
3 0

The percentage change in G is 21 %

<h3>What is Percentage change ?</h3>

Percentage change is defined as the increase or decrease in the value as compared to the original value multiplied by 100.

It is given that

G = ab

when a is increased by 10% the new a will be = 1.1 a

When b is increased by 10% the new b will be 1.1 b

So,

G' = 1.1a *1.1 b

G' = 1.21 ab

G' = 1.21

(G' - G)*100/G = (1.21-1)*100/1

The percentage change is 21 %

To know more about  percentage change

brainly.com/question/14979505

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Answer:

Yes

Step-by-step explanation:

It's just like adding fractions where you have to find the common denominator so the fractions can be subtracted.

<u>Example</u>

<u />\displaystyle \frac{1}{2}-\frac{1}{3}=\frac{1*3}{2*3}-\frac{2*1}{2*3}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}

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2 years ago
Simplify the expression: 14x-5(6x-7)+18​
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Answer:

−16x+53

Step-by-step explanation:

Distribute

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3 years ago
Tom has 2 more than 5 times the number of CD’s that Jane has. Jane has 5 CD’s. Write an
Soloha48 [4]

Step-by-step explanation:

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3 years ago
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If I make $14.00 an hour how much will I make monthly?
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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]

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6 0
3 years ago
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