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lubasha [3.4K]
3 years ago
7

In order to select new board members, the French club held an election. 50% of the 52 members of the club voted. How many member

s voted?
Mathematics
2 answers:
Lostsunrise [7]3 years ago
7 0

Answer:

26 members of the French club voted.

Step-by-step explanation:

<u>Find 50% of the 52 members that voted.</u>

50% · 52

50/100 · 52    

1/2 · 52

= 26      

kiruha [24]3 years ago
3 0

Answer:

26 Members Voted

Step-by-step explanation:

50 % is 50/100

Voted = (50/100) X 52 = 52/2 = 26

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If y varies directly as x and y = 21 when x = 5, find x when y = 42.
Lorico [155]

Answer: 10

<u>Step-by-step explanation:</u>

varies directly means: \frac{y}{x} = k

Step 1: solve for k   ⇒  \frac{21}{5} = k

Step 2: plug in the given value and k to solve for the missing value: \frac{42}{x} = \frac{21}{5}

42(5) = 21(x)

\frac{42(5)}{21} = \frac{21(x)}{21}

2(5) = x

10  = x


5 0
3 years ago
How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?
Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+...  \ \ \  \ \ --- (1)

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)          

f'(x) = \dfrac{1}{1+x}

f'(0)   = \dfrac{1}{1+0}=1

f ' ' (x)    = \dfrac{1}{(1+x)^2}

f ' ' (x)   = \dfrac{1}{(1+0)^2}=-1

f '  ' '(x)   = \dfrac{2}{(1+x)^3}

f '  ' '(x)    = \dfrac{2}{(1+0)^3} = 2

f ' '  ' '(x)    = \dfrac{6}{(1+x)^4}

f ' '  ' '(x)   = \dfrac{6}{(1+0)^4}=-6

f ' ' ' ' ' (x)    = \dfrac{24}{(1+x)^5} = 24

f ' ' ' ' ' (x)    = \dfrac{24}{(1+0)^5} = 24

Now, the next process is to substitute the above values back into equation (1)

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \  '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...

In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

To estimate the value of In(1.4), let's replace x with 0.4

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...

Therefore, from the above calculations, we will realize that the value of \dfrac{0.4^5}{5}= 0.002048 as well as \dfrac{0.4^6}{6}= 0.00068267 which are less than 0.001

Hence, the estimate of In(1.4) to the term is \dfrac{0.4^5}{5} is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0
3 years ago
Determine the numbers of solutions for 5x + 12 = 12x + 5.
mylen [45]

Answer:

Step-by-step explanation:

5x + 12 = 12x + 5

-7x + 12 = 5

-7x = -7

x = 1

one solution

8 0
3 years ago
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