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3 years ago

Is this question -true or false- I dont know

Mathematics

2 answers:

djyliett [7]3 years ago

8 0

Answer: FALSE

Step-by-step explanation:

kupik [55]3 years ago

6 0

The answer is false

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A mail order company charges 4.5% for shipping and handling. If the total order is 54.34 how much was the order total before shi

Andrews [41]

Answer:

$52

Step-by-step explanation:

Given that :

Amount charged for shipping and handling = 4.5%

Let cost of order = x

Total amount charged = 54.34

However,

total. Amount charged = (total cost + handling and shipping)

54.34 = x + 0.045x

54.34 = 1.045x

Order total before shipping = 54.34 / 1.045

Order total before shipping = 52

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3 years ago

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Effectus [21]

Answer:

Your answers are 1.A 2.T 3.C 4.A

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3 years ago

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2 years ago

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People use water to cook, clean, and drink every day. An estimate of 19.4% of the water used each day is for drinking. If a fa

Ronch [10]

Answer:

86.7 gallons

Step-by-step explanation:

I literally went decimal by decimal to find it. So theres no way it's wrong

4 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago